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Theory of Elasticity
Second Revised and Enlarged Edition
Course of Theoretical Physics
Volume 7
. D. Landau and E. M. Lifshi
Institute of Physical Problems
USSR Academy of Sciences
Pergamon Press
Course of Theoretical Physics
Volume 7
THEORY OF ELASTICITY
Second Revised and Enlarged Edition
L D. LANDAU and E. M. UFSHITZ
Institute of Physical Problems, USSR Academy
of Sciences
This is a comprehensive textbook covering
not only the ordinary theory of the
deformation of solids, but also some topics
not usually found in textbooks on the
subject, such as thermal conduction and
viscosity in solids, and various problems
in the theory of elastic vibrations and
waves. The authors have discussed only
briefly certain special matters, such as
complex mathematical methods in the
theory of elasticity and the theory of
shells, which are outside the scope of this
book.
As well as many minor corrections and
additions, a chapter on the macroscopic
theory of dislocations has been added in
this edition, while some further errors have
been eliminated.
COURSE OF THEORETICAL PHYSICS
Volume 7
THEORY OF ELASTICITY
OTHER TITLES IN THE SERIES
Vol. 1. MECHANICS
Vol. 2. THE CLASSICAL THEORY OF FIELDS
Vol. 3. QUANTUM MECHANICS— NON-RELATIVISTIC THEORY
Vol. 4. RELATIVISTIC QUANTUM THEORY
Vol. 5. STATISTICAL PHYSICS
Vol. 6. FLUID MECHANICS
Vol. 8. ELECTRODYNAMICS OF CONTINUOUS MEDIA
Vol. 9. PHYSICAL KINETICS
THEORY OF ELASTICITY
by
L. D. LANDAU and E. M. LIFSHITZ
INSTITUTE OF PHYSICAL PROBLEMS, U.S.S.R. ACADEMY OF SCIENCES
Volume 7 of Course of Theoretical Physics
Translated from the Russian by
J. B. SYKES and W. H. REID
SECOND ENGLISH EDITION, REVISED AND ENLARGED
PERGAMON PRESS
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All Rights Reserved. No part of this publication" may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any
means, electronic, mechanical, photocopying, recording or otherwise,
without the prior permission of Pergamon Press Limited.
First English edition 1959
Second (revised) edition 1970
Library of Congress Catalog Card No. 77-91701
Printed in Great Britain by J. W. Arrowsmith Ltd., Bristol
08 006465 5
CONTENTS
Page
Prefaces to the English editions vii
Notation viii
I. FUNDAMENTAL EQUATIONS
§1. The strain tensor 1
§2. The stress tensor 4
§3. The thermodynamics of deformation 8
§4. Hooke's law 10
§5. Homogeneous deformations 13
§6. Deformations with change of temperature 15
§7. The equations of equilibrium for isotropic bodies 17
§8. Equilibrium of an elastic medium bounded by a plane 25
§9. Solid bodies in contact 30
§10. The elastic properties of crystals 37
II. THE EQUILIBRIUM OF RODS AND PLATES
§11. The energy of a bent plate 44
§12. The equation of equilibrium for a plate 46
§13. Longitudinal deformations of plates 53
§14. Large deflections of plates 58
§15. Deformations of shells 62
§16. Torsion of rods 68
§17. Bending of rods 75
§18. The energy of a deformed rod 78
§19. The equations of equilibrium of rods 82
§20. Small deflections of rods 89
§21. The stability of elastic systems 97
III. ELASTIC WAVES
§22. Elastic waves in an isotropic medium 101
§23. Elastic waves in crystals 106
§24. Surface waves 109
§25. Vibration of rods and plates 113
§26. Anharmonic vibrations 118
IV. DISLOCATIONS
§27. Elastic deformations in the presence of a dislocation 123
§28. The action of a stress field on a dislocation 131
§29. A continuous distribution of dislocations 134
§30. Distribution of interacting dislocations 139
§31. Equilibrium of a crack in an elastic medium 144
vi Contents
V. THERMAL CONDUCTION AND VISCOSITY IN SOLIDS
Page
§32. The equation of thermal conduction in solids 150
§33. Thermal conduction in crystals 152
§34. Viscosity of solids 153
§35. The absorption of sound in solids 155
§36. Highly viscous fluids 161
Index 163
PREFACE TO THE FIRST ENGLISH EDITION
The present volume of our Theoretical Physics deals with the theory of
elasticity.
Being written by physicists, and primarily for physicists, it naturally
includes not only the ordinary theory of the deformation of solids, but also
some topics not usually found in textbooks on the subject, such as thermal
conduction and viscosity in solids, and various problems in the theory of
elastic vibrations and waves. On the other hand, we have discussed only
very briefly certain special matters, such as complex mathematical methods
in the theory of elasticity and the theory of shells, which are outside the scope
of this book.
Our thanks are due to Dr. Sykes and Dr. Reid for their excellent trans-
lation of the book.
Moscow L. D. Landau
E. M. Lifshitz
PREFACE TO THE SECOND ENGLISH EDITION
As well as some minor corrections and additions, a chapter on the macro-
scopic theory of dislocations has been added in this edition. The chapter has
been written jointly by myself and A. M. Kosevich.
A number of useful comments have been made by G. I. Barenblatt, V. L.
Ginzburg, M. A. Isakovich, I. M. Lifshitz and I. M. Shmushkevich for the
Russian edition, while the vigilance of Dr. Sykes and Dr. Reid has made it
possible to eliminate some further errors from the English translation.
I should like to express here my sincere gratitude to all the above-named.
Moscow E. M. Lifshitz
NOTATION
p density of matter
u displacement vector
1 / dut duiA
uuc = -I 1 1 strain tensor
2\dxjc dxi
aoe stress tensor
K modulus of compression
/x modulus of rigidity
E Young's modulus
a Poisson's ratio
c% longitudinal velocity of sound
ct transverse velocity of sound
Ci and Ct are expressed in terms of K, ju, or of E, a by formulae given in
§22.
The quantities K, p, E and a are related by
E = 9KfMl(3K+fi)
a = (3K-2fM)l2(3K+tJL)
K= EI3(l-2o)
fi = £"/2(l+a)
CHAPTER I
FUNDAMENTAL EQUATIONS
§1. The strain tensor
The mechanics of solid bodies, regarded as continuous media, forms the
content of the theory of elasticity.]
Under the action of applied forces, solid bodies exhibit deformation to
some extent, i.e. they change in shape and volume. The deformation of a
body is described mathematically in the following way. The position of any
point in the body is defined by its radius vector r (with components x\ = x,
x 2 = y, xz = z) in some co-ordinate system. When the body is deformed,
every point in it is in general displaced. Let us consider some particular
point; let its radius vector before the deformation be r, and after the deforma-
tion have a different value r' (with components x'i). The displacement of
this point due to the deformation is then given by the vector r' - r, which we
shall denote by u:
Ui = x'i — Xi. (1-1)
The vector u is called the displacement vector. The co-ordinates x'i of the
displaced point are, of course, functions of the co-ordinates Xi of the point
before displacement. The displacement vector Ui is therefore also a function
of the co-ordinates Xi. If the vector u is given as a function of Xi, the defor-
mation of the body is entirely determined.
When a body is deformed, the distances between its points change. Let
us consider two points very close together. If the radius vector joining them
before the deformation is dxt, the radius vector joining the same two points
in the deformed body is dx'i = dxt + dm. The distance between the points
is d/ = -v/(d*i 2 + d#2 2 + dff 3 2 ) before the deformation, and dl' = \/(dx'i 2 +
+ dx'2 2 +dx's 2 ) after it. Using the general summation rule, J we can write
d/ 2 = dxi 2 , dl' 2 = dx'i 2 = (dxi + dui) 2 . Substituting dm = (dmldx k )dx k ,we
can write
dm . . dm dm ,
dl' 2 = d/ 2 + 2 dxt dx k + dx k dxi.
dx k dx k dxi
Since the summation is taken over both suffixes i and k in the second term
on the right, we can put (dmldx k )dxidx k = {du k jdxi)dxidx k . In the third
t The basic equations of elasticity theory were established in the 1 820's by Cauchy and by Poisson.
| In accordance with the usual rule, we omit the sign of summation over vector and tensor suffixes.
Summation over the values 1 , 2, 3 is understood with respect to all suffixes which appear twice in a
given term.
I* 1
2 Fundamental Equations §1
term, we interchange the suffixes i and /. Then dZ' 2 takes the final form
d/'2 = dP + 2u ik dxt dx k , (1.2)
where the tensor ut k is defined as
1 / dui dujc dui dui \
2\dx k dxt dxt dxjc)
These expressions give the change in an element of length when the body is
deformed.
The tensor u\ k is called the strain tensor. We see from its definition that
it is symmetrical, i.e.
Uik = u ki . (1.4)
This result has been obtained by writing the term 2(dui/dx k )dxi dx k in d/' 2
in the explicitly symmetrical form
/ But du k \
\ dx k oxi I
Like any symmetrical tensor, Uf k can be diagonalised at any given point.
This means that, at any given point, we can choose co-ordinate axes (the
principal axes of the tensor) in such a way that only the diagonal components
«n, "22, «33 of the tensor u\ k are different from zero. These components, the
principal values of the strain tensor, will be denoted by m (1) , m (2) , m (3) . It should
be remembered, of course, that, if the tensor ui k is diagonalised at any point
in the body, it will not in general be diagonal at any other point.
If the strain tensor is diagonalised at a given point, the element of length
(1.2) near it becomes
dZ' 2 = (Sac + 2uuc) dxt dx k
= (1 + 2 M <i>) d*i 2 + (1 + 2w< 2 >) d*2 2 + (1 + 2m«») dx 3 2 .
We see that the expression is the sum of three independent terms. This
means that the strain in any volume element may be regarded as composed
of independent strains in three mutually perpendicular directions, namely
those of the principal axes of the strain tensor. Each of these strains is a
simple extension (or compression) in the corresponding direction : the length
d#i along the first principal axis becomes dx\ = -\/(l+2u a) ) dx\, and simi-
larly for the other two axes. The quantity \/{\ + 2u (i) ) — 1 is consequently
equal to the relative extension (dx'i — dxi)[dxt along the ith principal axis.
In almost all cases occurring in practice, the strains are small. This means
that the change in any distance in the body is small compared with the
distance itself. In other words, the relative extensions are small compared
with unity. In what follows we shall suppose that all strains are small.
If a body is subjected to a small deformation, all the components of the
strain tensor are small, since they give, as we have seen, the relative changes
in lengths in the body. The displacement vector Ui, however, may
sometimes be large, even for small strains. For example, let us consider a
long thin rod. Even for a large deflection, in which the ends of the rod move
§1 The strain tensor 3
a considerable distance, the extensions and compressions in the rod itself
will be small.
Except in such special cases, f the displacement vector for a small defor-
mation is itself small. For it is evident that a three-dimensional body (i.e.
one whose dimension in no direction is small) cannot be deformed in such a
way that parts of it move a considerable distance without the occurrence of
considerable extensions and compressions in the body.
Thin rods will be discussed in Chapter II. In other cases m is small for
small deformations, and we can therefore neglect the last term in the general
expression (1.3), as being of the second order of smallness. Thus, for small
deformations, the strain tensor is given by
_ 1 tM duk\
2\dxic dxil
The relative extensions of the elements of length along the principal axes of
the strain tensor (at a given point) are, to within higher-order quantities,
VXl +2u (i) )— 1 & u (i) , i.e. they are the principal values of the tensor tin.
Let us consider an infinitesimal volume element dV, and find its volume
dV after the deformation. To do so, we take the principal axes of the strain
tensor, at the point considered, as the co-ordinate axes. Then the elements of
length d^i, d*2, dx$ along these axes become, after the deformation, d#'i
= (l+w (1) ) d#i, etc. The volume dV is the product d#i da^ d#3, while dV'
is dx\ dx' 2 dx' z . Thus dV = dV{\ + u<-»)(l + w (2) )(l + m (3) ). Neglecting higher-
order terms, we therefore have dV = dF(l+« (1) + M (2) + M (3) ). The sum
M (D -|- w (2) + M (3) of the principal values of a tensor is well known to be invariant,
and is equal to the sum of the diagonal components uu =Mn + M22+«33 in
any co-ordinate system. Thus
dV = dV(l + u u ). (1.6)
We see that the sum of the diagonal components of the strain tensor is the
relative volume change {dV — dV)jdV.
It is often convenient to use the components of the strain tensor in spherical
or cylindrical co-ordinates. We give here, for reference, the corresponding
formulae, which express the components in terms of the derivatives of the
components of the displacement vector in the same co-ordinates. In spherical
co-ordinates r, 6, (f>, we have
du r 1 du e u r 1 du* u e u r
u r r = —, u ee = -—-+—-, u H = —^ -— - + — cot0 + — ,
dr r do r r sin o of r r
Y r\dd Y J r si]
dUg BUg Ug 1 Bu r > (1.7)
2u r e = — +-■
sin0 9^' dr r r dd
1 du r dUs, Ux
2ufr = H -•
r sin 9 d<f> dr r
t Which include, besides deformations of thin rods, those of thin plates to form cylindrical surfaces.
4 Fundamental Equations §2
In cylindrical co-ordinates r, (f>, z,
du r 1 duf u r du z
u rr = -r— , «^ = - —7- H , u zz = — — ,
or yr r d<f> r dz
1 du z du A du r du z
2«„ = -—+—*, 2u rz = -^ + -^- f } (1.8)
r 30 d# 3s- 3r
aw, W, 1 du r
or r r dtp
1
§2. The stress tensor
In a body that is not deformed, the arrangement of the molecules corre-
sponds to a state of thermal equilibrium. All parts of the body are in mechani-
cal equilibrium. This means that, if some portion of the body is considered,
the resultant of the forces on that portion is zero.
When a deformation occurs, the arrangement of the molecules is changed,
and the body ceases to be in its original state of equilibrium. Forces there-
fore arise which tend to return the body to equilibrium. These internal
forces which occur when a body is deformed are called internal stresses. If
no deformation occurs, there are no internal stresses.
The internal stresses are due to molecular forces, i.e. the forces of inter-
action between the molecules. An important fact in the theory of elasticity is
that the molecular forces have a very short range of action. Their effect
extends only to the neighbourhood of the molecule exerting them, over a
distance of the same order as that between the molecules, whereas in the
theory of elasticity, which is a macroscopic theory, the only distances con-
sidered are those large compared with the distances between the molecules.
The range of action of the molecular forces should therefore be taken as zero
in the theory of elasticity. We can say that the forces which cause the internal
stresses are, as regards the theory of elasticity, "near-action" forces, which act
from any point only to neighbouring points. Hence it follows that the forces
exerted on any part of the body by surrounding parts act only on the surface
of that part.
The following reservation should be made here. The above asserioon is
not valid in cases where the deformation of the body results in macroscopic
electric fields in it (pyroelectric and piezoelectric bodies). We shall not discuss
such bodies in this book, however.
Let us consider the total force on some portion of the body. Firstly, this
total force is equal to the sum of all the forces on all the volume elements in
that portion of the body, i.e. it can be written as the volume integral jTdV,
where F is the force per unit volume and FdFthe force on the volume element
dV. Secondly, the forces with which various parts of the portion considered
act on one another cannot give anything but zero in the total resultant force,
since they cancel by Newton's third law. The required total force can there-
fore be regarded as the sum of the forces exerted on the given portion of the
§2 The stress tensor 5
body by the portions surrounding it. From above, however, these forces act
on the surface of that portion, and so the resultant force can be represented
as the sum of forces acting on all the surface elements, i.e. as an integral
over the surface.
Thus, for any portion of the body, each of the three components j FfdV
of the resultant of all the internal stresses can be transformed into an integral
over the surface. As we know from vector analysis, the integral of a scalar
over an arbitrary volume can be transformed into an integral over the surface
if the scalar is the divergence of a vector. In the present case we have the
integral of a vector, and not of a scalar. Hence the vector Ft must be the
divergence of a tensor of rank two, i.e. be of the form
Ft = doac/dxjc. (2.1)
Then the force on any volume can be written as an integral over the closed
surface bounding that volume :f
IV, dV = f ^ dV = L ik d/fc, (2.2)
where d/< are the components of the surface element vector df, directed (as
usual) along the outward normal. J
The tensor aw is called the stress tensor. As we see from (2.2), aacdfjc is the
tth component of the force on the surface element df. By taking elements
of area in the planes of xy, yz, zx, we find that the component am of the stress
tensor is the ith. component of the force on unit area perpendicular to the
x*-axis. For instance, the force on unit area perpendicular to the ar-axis,
normal to the area (i.e. along the #-axis) is <t xx , and the tangential forces
(along the y and z axes) are a yx and a zx .
The following remark should be made concerning the sign of the force
oikdfk. The surface integral in (2.2) is the force exerted on the volume
enclosed by the surface by the surrounding parts of the body. The force
which this volume exerts on the surface surrounding it is the same with the
opposite sign. Hence, for example, the force exerted by the internal stresses
on the surface of the body itself is —fontdfk, where the integral is taken over
the surface of the body and df is along the outward normal.
Let us determine the moment of the forces on a portion of the body. The
moment of the force F can be written as an antisymmetrical tensor of rank
two, whose components are FiXk — FjcXi, where xi are the co-ordinates of the
t The integral over a closed surface is transformed into one over the volume enclosed by the
surface by replacing the surface element d/ 4 by the operator dVd/dxi.
J Strictly speaking, to determine the total force on a deformed portion of the body we should
integrate, not over the old co-ordinates x t , but over the co-ordinates x' t of the points of the deformed
body. The derivatives (2.1) should therefore be taken with respect to x' t . However, in view of the
smallness of the deformation, the derivatives with respect to x ( and x\ differ only by higher-order
quantities, and so the derivatives can be taken with respect to the co-ordinates x t .
6 Fundamental Equations §2
point where the force is applied.f Hence the moment of the forces on the
volume element dV is (FiX k — F k Xi)dV, and the moment of the forces on the
whole volume is Mi k = J(FiX k — F k Xi)dV. Like the total force on any volume,
this moment can be expressed as an integral over the surface bounding the
volume. Substituting the expression (2.1) for Fi, we find
M ik = (— — x k — xA dV
J \ dxi oxi ]
rd(auXjc-o k iXi) C( dx k dx t \
= ; dv ~ au i> ° kl v~ dv -
J oxi J \ dxi dxi !
In the second term we use the fact that the derivative of a co-ordinate with
respect to itself is unity, and with respect to another co-ordinate is zero
(since the three co-ordinates are independent variables). Thus dx k jdxi = 8 k i,
where Sfcz is the unit tensor; the multiplication gives CT^Sfc/ = at k , a k ion = <?ki-
In the first term, the integrand is the divergence of a tensor; the
integral can be transformed into one over the surface. The result is
Mac = §(cTux k - ajciXi)dfi+ S(crjci - o ik )dV. If M ik is to be an integral over the
surface only, the second term must vanish identically, i.e. we must have
&ik = <*ki- (2.3)
Thus we reach the important result that the stress tensor is symmetrical.
The moment of the forces on a portion of the body can then be written
simply as
M ik = j(FiX k - F k x t ) d V = j((T U x k - a k iXi) dfi . (2.4)
It is easy to find the stress tensor for a body undergoing uniform com-
pression from all sides {hydrostatic compression). In this case a pressure of
the same magnitude acts on every unit area on the surface of the body, and its
direction is along the inward normal. If this pressure is denoted by />, a force
—pdfi acts on the surface element dfi. This force, in terms of the stress
tensor, must be cxi k df k . Writing — pdfi = —p8i k df k , we see that the stress
tensor in hydrostatic compression is
f^ik = -poik- (2.5)
Its non-zero components are simply equal to the pressure.
In the general case of an arbitrary deformation, the non-diagonal com-
ponents of the stress tensor are also non-zero. This means that not only a
normal force but also tangential (shearing) stresses act on each surface
element. These latter stresses tend to move the surface elements relative to
each other.
f The moment of the force F is denned as the vector product FXr, and we know from vector
analysis that the components of a vector product form an antisymmetrical tensor of rank two as written
here.
§2 The stress tensor 1
In equilibrium the internal stresses in every volume element must balance,
i.e. we must have Ft = 0. Thus the equations of equilibrium for a deformed
body are
dcTiicjdxjc = 0. (2.6)
If the body is in a gravitational field, the sum F + />g of the internal stresses
and the force of gravity (pg per unit volume) must vanish ; p is the density f and
g the gravitational acceleration vector, directed vertically downwards. In
this case the equations of equilibrium are
daikldx k +pgi = 0. (2.7)
The external forces applied to the surface of the body (which are the usual
cause of deformation) appear in the boundary conditions on the equations of
equilibrium. Let P be the external force on unit area of the surface of the
body, so that a force P d/ acts on a surface element d/. In equilibrium, this
must be balanced by the force — aw d/& of the internal stresses acting on that
element. Thus we must have Pi df— a i1c df k = 0. Writing d/fc = tik df,
where n is a unit vector along the outward normal to the surface, we find
(Wit = Pi. (2.8)
This is the condition which must be satisfied at every point on the surface of
a body in equilibrium.
We shall derive also a formula giving the mean value of the stress tensor
in a deformed body. To do so, we multiply equation (2.6) by xjc and integrate
over the whole volume:
——xjc dV = -\ - dV- an—- dV = 0.
J oxi J oxi J oxi
The first integral on the right is transformed into a surface integral; in the
second integral we put dxjc/dxi = 8m. The result is §auxjc d/i — jaw dV = 0.
Substituting (2.8) in the first integral, we find §PiXjc df = JV^ dV = Vdnc,
where V is the volume of the body and d^ the mean value of the stress tensor.
Since a^ = ajd, this formula can be written in the symmetrical form
a m = (1/2F) j (PiXk + P k xi) df. (2. 9)
Thus the mean value of the stress tensor can be found immediately from the
external forces acting on the body, without solving the equations of equili-
brium.
t Strictly speaking, the density of a body changes when it is deformed. An allowance for this
change, however, involves higher-order quantities in the case of small deformations, and is therefore
unimportant.
8 Fundamental Equations §3
§3. The thermodynamics of deformation
Let us consider some deformed body, and suppose that the deformation
is changed in such a way that the displacement vector m changes by a small
amount dm; and let us determine the work done by the internal stresses in
this change. Multiplying the force Fi = dcrikldxjc by the displacement Sw$ and
integrating over the volume of the body, we have J8R dV = ftdaikjdxkjSui dV,
where 8R denotes the work done by the internal stresses per unit volume.
We integrate by parts, obtaining
8RdV = koikhuidfjc— <*ik— — dV.
By considering an infinite medium which is not deformed at infinity, we
make the surface of integration in the first integral tend to infinity; then
aijc = on the surface, and the integral is zero. The second integral can,
by virtue of the symmetry of the tensor cm, be written
C 1 r I dhui dhujA
J Z.J \ UXfc UA% J
- _if hi— dUk \ dV
2 J \dxjc dxi/
= — vik&Uik dV.
Thus we find
SR = —oikhuoc. (3.1)
This formula gives the work SR in terms of the change in the strain tensor.
If the deformation of the body is fairly small, it returns to its original
undeformed state when the external forces causing the deformation cease
to act. Such deformations are said to be elastic. For large deformations, the
removal of the external forces does not result in the total disappearance of the
deformation ; a residual deformation remains, so that the state of the body is
not that which existed before the forces were applied. Such deformations
are said to be plastic. In what follows we shall consider only elastic defor-
mations.
We shall also suppose that the process of deformation occurs so slowly
that thermodynamic equilibrium is established in the body at every instant,
in accordance with the external conditions. This assumption is almost always
justified in practice. The process will then be thermodynamically reversible.
In what follows we shall take all such thermodynamic quantities as the
entropy S, the internal energy $, etc., relative to unit volume of the body,f
f The following remark should be made here. Strictly speaking, the unit volumes before and after
the deformation should be distinguished, since they in general contain different amounts of matter.
We shall always relate the thermodynamic quantities to unit volume of the undeformed body, i.e.
to the amount of matter therein, which may occupy a different volume after the deformation. Accord-
ingly, the total energy of the body, for example, is obtained by integrating £ over the volume of the
undeformed body.
§3 The thermodynamics of deformation 9
and not relative to unit mass as in fluid mechanics, and denote them by the
corresponding capital letters.
An infinitesimal change d& in the internal energy is equal to the difference
between the heat acquired by the unit volume considered and the work dR
done by the internal stresses. The amount of heat is, for a reversible process,
TdS, where T is the temperature. Thus d£ = TdS-dR; with dR given
by (3.1), we obtain
d£ = TdS+a ik du ik . (3.2)
This is the fundamental thermodynamic relation for deformed bodies.
In hydrostatic compression, the stress tensor is a ik = —phk (2.5). Then
one duac = -phk dune = -p duu. We have seen, however (cf. (1.6)), that the
sum uu is the relative volume change due to the deformation. If we consider
unit volume, therefore, uu is simply the change in that volume, and dun is
the volume element dV. The thermodynamic relation then takes its usual form
te = TdS-pdV.
Introducing the free energy of the body, F = S— TS, we find the form
dF = -SdT+a ik du ik (3.3)
of the relation (3.2). Finally, the thermodynamic potential $ is defined as
<J> = £-TS-o ik u ik = F-a ik u ik . (3.4)
This is a generalisation of the usual expression O = <f — TS+pV.-f Substi-
tuting (3.4) in (3.3), we find
dd> = -SdT-u ik d(j ik . (3.5)
The independent variables in (3.2) and (3.3) are respectively S, u ik and
T, Ui k . The components of the stress tensor can be obtained by differentiating
S or F with respect to the components of the strain tensor, for constant
entropy S or temperature T respectively:
a ik = (d<?ldu ik )s = {dFjdu ik ) T - (3.6)
Similarly, by differentiating <1> with respect to the components <j ik , we can
obtain the components u\ k :
u ik = -(d^lda ik ) T . (3.7)
t For hydrostatic compression, the expression (3.4) becomes <b= F+ pu it = F + p(V —V ),
where V —V is the volume change resulting from the deformation. Hence we see that the definition
of used here differs by a term — pV from the usual definition O = F+ pV.
10 Fundamental Equations §4
§4. Hooke's law
In order to be able to apply the general formulae of thermodynamics to
any particular case, we must know the free energy F of the body as a function
of the strain tensor. This expression is easily obtained by using the fact that
the deformation is small and expanding the free energy in powers of w^Jfc- We
shall at present consider only isotropic bodies. The corresponding results
for crystals will be obtained in §10.
In considering a deformed body at some temperature (constant throughout
the body), we shall take the undeformed state to be the state of the body in the
absence of external forces and at the same temperature ; this last condition is
necessary on account of the thermal expansion (see §6). Then, for ua = 0,
the internal stresses are zero also, i.e. a^ = 0. Since ow = dFjdutk, it
follows that there is no linear term in the expansion of F in powers of uac.
Next, since the free energy is a scalar, each term in the expansion of F
must be a scalar also. Two independent scalars of the second degree can be
formed from the components of the symmetrical tensor m^: they can be
taken as the squared sum of the diagonal components (uu 2 ) and the sum of
the squares of all the components (uac 2 ). Expanding F in powers of «^» we
therefore have as far as terms of the second order
F = Fo+%\u u 2 + iJLU ik 2 . (4.1)
This is the general expression for the free energy of a deformed isotropic
body. The quantities A and [m are called Lame coefficients.
We have seen in §1 that the change in volume in the deformation is given
by the sum uu. If this sum is zero, then the volume of the body is unchanged
by the deformation, only its shape being altered. Such a deformation is
called a pure shear.
The opposite case is that of a deformation which causes a change in the
volume of the body but no change in its shape. Each volume element of the
body retains its shape also. We have seen in §1 that the tensor of such a
deformation is uac = constant x S^. Such a deformation is called a hydro-
static compression.
Any deformation can be represented as the sum of a pure shear and a
hydrostatic compression. To do so, we need only use the identity
mk = (uuc-$8ikUii) + %8 ik uu. (4.2)
The first term on the right is evidently a pure shear, since the sum of its
diagonal terms is zero (8u = 3). The second term is a hydrostatic compres-
sion.
As a general expression for the free energy of a deformed isotropic body,
it is convenient to replace (4.1) by another formula, using this decomposition
of an arbitrary deformation into a pure shear and a hydrostatic compression.
We take as the two independent scalars of the second degree the sums of the
§4 Hooke's law 1 1
squared components of the two terms in (4.2). Then F becomesf
F = 11(11* - \hmif + Wm*. (4.3)
The quantities K and ju. are called respectively the bulk modulus or modulus of
hydrostatic compression (or simply the modulus of compression) and the shear
modulus or modulus of rigidity. K is related to the Lame coefficients by
K = A + foi. (4.4)
In a state of thermodynamic equilibrium, the free energy is a minimum.
If no external forces act on the body, then F as a function of uw must have a
minimum for uac - 0. This means that the quadratic form (4.3) must be
positive. If the tensor u ilc is such that uu = 0, only the first term remains
in (4.3) ; if, on the other hand, the tensor is of the form u* = constant x S^,
then only the second term remains. Hence it follows that a necessary (and
evidently sufficient) condition for the form (4.3) to be positive is that each
of the coefficients K and ju, is positive. Thus we conclude that the moduli of
compression and rigidity are always positive:
K > 0, ft > 0. (4.5)
We now use the general thermodynamic relation (3.6) to determine the
stress tensor. To calculate the derivatives dFjduijc, we write the total differ-
ential dF (for constant temperature) :
dF = Kuu dun + 2p{uui - \uiihoc) d(uijc - Iuu^m).
In the second term, multiplication of the first parenthesis by oik gives zero,
leaving dF = Ku u duii + 2[4iHk-%uu8ik) &Uik, or writing dun = 8 ik duac,
dF = [Kuuhuc + l^uuc-^uiihijc)] du ik .
Hence the stress tensor is
G ilc = Kuuhi]c + 2^{uiic-\8iicUii). (4.6)
This expression determines the stress tensor in terms of the strain tensor for
an isotropic body. It shows, in particular, that, if the deformation is a pure
shear or a pure hydrostatic compression, the relation between a* and uac is
determined only by the modulus of rigidity or of hydrostatic compression
respectively.
It is not difficult to obtain the converse formula which expresses uac in
terms of 0%. To do so, we find the sum an of the diagonal terms. Since this
sum is zero for the second term of (4.6), we have an = 3Kuu, or
uu = oiifiK. (4.7)
f The constant term F is the free energy of the undeformed body, and is of no further interest.
We shall therefore omit it, for brevity, taking F to be only the free energy of the deformation (the
elastic free energy, as it is called).
12 Fundamental Equations §4
Substituting this expression in (4.6) and so determining u ik , we find
utjc = S ik aiij9K+(a ik -l8 ik aii)/2fi y (4.8)
which gives the strain tensor in terms of the stress tensor.
Equation (4.7) shows that the relative change in volume (uu) in any
deformation of an isotropic body depends only on the sum a u of the diagonal
components of the stress tensor, and the relation between uu and <r u is
determined only by the modulus of hydrostatic compression. In hydrostatic
compression of a body, the stress tensor is a ik = -p8 ik . Hence we have
in this case, from (4.7),
uu = -p/K. (4.9)
Since the deformations are small, uu and p are small quantities, and we can
write the ratio uu/p of the relative volume change to the pressure in the
differential form (l/V)(dVfdp) T . Thus
JL_ 1 ( dV \
~K~ ~ V\dp) t
The quantity \\K is called the coefficient of hydrostatic compression (or simply
the coefficient of compression).
We see from (4.8) that the strain tensor Ui k is a linear function of the stress
tensor Oi k . That is, the deformation is proportional to the applied forces.
This law, valid for small deformations, is called Hooke's law.f
We may give also a useful form of the expression for the free energy of a
deformed body, which is obtained immediately from the fact that F is quad-
ratic in the strain tensor. According to Euler's theorem, Ui k dFjdui k — 2F,
whence, since dF\dui k = oi ki we have
F = \o ik u i1c . (4.10)
If we substitute in this formula the ua as linear combinations of the
components oi ky the elastic energy will be represented as a quadratic function
of the ai k . Again applying Euler's theorem, we obtain cr ik dF/da ik = 2F, and
a comparison with (4.10) shows that
u ik = dFjda ik . (4.11)
It should be emphasised, however, that, whereas the formula o% = dF/duac
is a general relation of thermodynamics, the inverse formula (4.11) is applic-
able only if Hooke's law is valid.
f Hooke's law is actually applicable to almost all elastic deformation. The reason is that deforma-
tions usually cease to be elastic when they are still so small that Hooke's law is a good approximation.
Substances such as rubber form an exception.
§5 Homogeneous deformations 13
§5. Homogeneous deformations
Let us consider some simple cases of what are called homogeneous deforma-
tions, i.e. those in which the strain tensor is constant throughout the volume
of the body. For example, the hydrostatic compression already considered
is a homogeneous deformation.
We first consider a simple extension (or compression) of a rod. Let the
rod be along the #-axis, and let forces be applied to its ends which stretch it
in both directions. These forces act uniformly over the end surfaces of the
rod ; let the force on unit area be p.
Since the deformation is homogeneous, i.e. uuc is constant through the
body, the stress tensor o% is also constant, and so it can be determined at once
from the boundary conditions (2.8). There is no external force on the sides
of the rod, and therefore o-^« & = 0. Since the unit vector n on the side of the
rod is perpendicular to the #-axis, i.e. n z = 0, it follows that all the com-
ponents aw except o zz are zero. On the end surface we have a z tni = p, or
ozz = P-
From the general expression (4.8) which relates the components of the
strain and stress tensors, we see that all the components uuc with i ^ k are
zero. For the remaining components we find
1/1 1 \ 1/ 1 1\
u xx = u yy - --(_ - _jp, u zz = 3(3^ + -)* (5-1)
The component u zz gives the relative lengthening of the rod. The coeffi-
cient of p is called the coefficient of extension, and its reciprocal is the modulus
of extension or Young's modulus, E:
u zz = pJE, (5.2)
where
E = 9Kfjil(3K+fji). (5.3)
The components u X x and u yy give the relative compression of the rod in the
transverse direction. The ratio of the transverse compression to the longi-
tudinal extension is called Poisson's ratio, cr:j-
u xx = -<yu zz , (5.4)
where
o = %(3K-2 H ,)/(3K+ H ,). (5.5)
f The use of a to denote Poisson's ratio and a^ to denote the components of the stress tensor can-
not lead to ambiguity, since the latter always have suffixes.
14
Fundamental Equations
§5
Since K and p are always positive, Poisson's ratio can vary between - 1
(for K = 0) and J (for p = 0). Thus f
- 1 < <t < i. (5.6)
Finally, the relative increase in the volume of the rod is
m = pllK. (5.7)
The free energy of a stretched rod can be obtained immediately from formula
(4.10). Since only the component a zz is not zero, we have F = \a zz u zz ,
whence
F = p2/2E. (5.8)
In what follows we shall, as is customary, use E and a instead of K and p.
Inverting formulae (5.3) and (5.5), we havej
p = E/2(l + cx), K = 5/3(1 -2a). (5.9)
We shall write out here the general formulae of §4, with the coefficients
expressed in terms of E and a. The free energy is
y -^ + i^)' (5 - 10)
The stress tensor is given in terms of the strain tensor by
E t a
Vik
Conversely,
1 + a
Uik +
l-2a
mS
ik\
Uik = [( 1 + a ) aw - oaiiSuc] IE.
(5.11)
(5.12)
Since formulae (5.11) and (5.12) are in frequent use, we shall give them also
in component form :
E \
&xx = tt - — rz — ^r-z[(l- a ) u xx + o(uyy + u zz )],
f yy
&ZZ
o X y
(l+a)(l-2a) u
E
(l+a)(l-2a)
E
(l+a)(l-2a)
E
Uxy, &xz =
[(1 - a)u yy + a(u xx + U zz )],
[(1 - a)U ZZ + 0(U XX + Uyy)],
E
E
1+CT
1 + a
u xtt a yz
1+a
Uy Z ,
(5.13)
t In practice, PoiSSON's ratio varies only between and i. There are no substances known for
which a < 0, i.e. which would expand transversely when stretched longitudinally. It may be men-
tioned that the inequality a > corresponds to A > 0, where A is the Lame coefficient appearing
in (4.1); in other words, both terms in (4.1), as well as in (4.3), are always positive in practice, although
this is not thermodynamically necessary. Values of a close to i (e.g. for rubber) correspond to a
modulus of rigidity which is small compared with the modulus of compression.
t The second Lame coefficient is A = Eaj{\ — 2a)(14-a).
6 Deformations with change of temperature
and conversely
1
U X z = — \?xx— o - (o'j/2/ + o'zz)].
u vv — ~^;\. a yy~ < T ( cr zx+ &zz)]>
hi
15
U zz = —[cr Z z — cr(oxx+Vyy)]>
hi
( (5-14)
U X y
1 + a
E
-a X y, U X z —
1 + a
E
-<7xZt Uy Z —
1 + a
~E
~ a yz'
Let us now consider the compression of a rod whose sides are fixed in
such a way that they cannot move. The external forces which cause the
compression of the rod are applied to its ends and act along its length, which
we again take to be along the #-axis. Such a deformation is called a
unilateral compression. Since the rod is deformed only in the ^-direction,
only the component u zz of m^ is not zero. Then we have from (5.11)
E
o X x — a yy —
(l + a)(l-2cr)
U ZZ> &ZZ —
27(1 -Or)
(l + o)(l-2o)
u zz .
Again denoting the compressing force by p (a zz = p, which is negative for
a compression), we have
Uzz = />(1 + a)(l - 2a)lE(l - a).
(5.15)
The coefficient of p is called the coefficient of unilateral compression. For the
transverse stresses we have
°xx = Oyy = />a/(l — o).
Finally, the free energy of the rod is
F = />2(1 + CT )(l _ 2a)/2E(l - a).
(5.16)
(5.17)
§6. Deformations with change of temperature
Let us now consider deformations which are accompanied by a change in
the temperature of the body; this can occur either as a result of the deforma-
tion process itself, or from external causes.
We shall regard as the undeformed state the state of the body in the absence
of external forces at some given temperature To. If the body is at a tempera-
ture T different from To, then, even if there are no external forces, it will in
general be deformed, on account of thermal expansion. In the expansion of
the free energy F(T), there will therefore be terms linear, as well as quadratic,
in the strain tensor. From the components of the tensor uac, of rank two,
16 Fundamental Equations §6
we can form only one linear scalar quantity, the sum uu of its diagonal com-
ponents. We shall also assume that the temperature change T—Tq which
accompanies the deformation is small. We can then suppose that the coeffi-
cient of uu in the expansion of F (which must vanish for T = To) is simply
proportional to the difference T— To. Thus we find the free energy to be
(instead of (4.3))
F(T) = Fo{T)-K*{T- Toyn+niuijc-^ncUnY + Wm 2 , (6.1)
where the coefficient of T—Tq has been written as — Ktx.. The quantities
/Lt, K and a can here be supposed constant; an allowance for their tempera-
ture dependence would lead to terms of higher order.
Differentiating F with respect to um, we obtain the stress tensor:
a ik = -KatT-To^iK + KunSiit + lrfuik-iSiKUH). (6.2)
The first term gives the additional stresses caused by the change in tempera-
ture. In free thermal expansion of the body (external forces being absent),
there can be no internal stresses. Equating a^ to zero, we find that ui k is of
the form constant x §*&, and
uu = <x(T— T ). (6.3)
But uu is the relative change in volume caused by the deformation. Thus a
is just the thermal expansion coefficient of the body.
Among the various (thermodynamic) types of deformation, isothermal and
adiabatic deformations are of importance. In isothermal deformations, the
temperature of the body does not change. Accordingly, we must put T = To
in (6.1), returning to the usual formulae; the coefficients K and //, may there-
fore be called isothermal moduli.
A deformation is adiabatic if there is no exchange of heat between the
various parts of the body (or, of course, between the body and the surround-
ing medium). The entropy S remains constant. It is the derivative — dFjdT
of the free energy with respect to temperature. Differentiating the expression
(6.1), we have as far as terms of the first order in utk
S(T) = S {T) + Kauu. (6.4)
Putting S constant, we can determine the change of temperature T—T due
to the deformation, which is therefore proportional to uu. Substituting this
expression for T- T in (6.2), we obtain for a ik an expression of the usual
kind,
oik = Ka,aUiiSik + 2[j,(uik-^SikUii) t (6.5)
with the same modulus of rigidity \x. but a different modulus of compression
Kad. The relation between the adiabatic modulus K &A and the ordinary
isothermal modulus K can also be found directly from the thermodynamic
formula
/ dV\ _ (W\ T(dVldT)j?
\ dp i s \ dp / t C
p
§7 The equations of equilibrium for isotropic bodies 17
where C p is the specific heat per unit volume at constant pressure. If V is
taken to be the volume occupied by matter which before the deformation
occupied unit volume, the derivatives dVjdT and dVjdp give the relative
volume changes in heating and compression respectively. That is,
(dV/dT) p = a, (dVldp) s = -1/JT.d, (dVldp) T = -1(K.
Thus we find the relation between the adiabatic and isothermal moduli to be
l/JSTad = 1/K- TaPICp, juad = H- (6-6)
For the adiabatic Young's modulus and Poisson's ratio we easily obtain
E o + ET**I9C p
E& * = \-ET^\9C p Cad " 1-ET«*I9C P - {bJ)
In practice, ETa. 2 jC p is usually small, and it is therefore sufficiently accurate
to put
£ad = E+ E2T<x?j9C p , cr ad = a + (1 + o)ETa?l9C p . (6.8)
In isothermal deformation, the stress tensor is given in terms of the
derivatives of the free energy :
aik = (dF/duacJT'
For constant entropy, on the other hand, we have (see (3.6))
aw = {d$lduac)s,
where & is the internal energy. Accordingly, the expression analogous to
(4.3) determines, for adiabatic deformations, not the free energy but the in-
ternal energy per unit volume :
£ = \K & mi 2 + K U M ~ \mhkf. (6.9)
§7. The equations of equilibrium for isotropic bodies
Let us now derive the equations of equilibrium for isotropic solid bodies.
To do so, we substitute in the general equations (2.7)
doijc/dxjc + pgi =
the expression (5.11) for the stress tensor. We have
daw Ea dun E duik
+ ■
dxic (1 + ct)(1-2ct) dxi 1 + ct dxic
Substituting
l / dut du]c\
2\dx]c dx t i
18 Fundamental Equations §7
we obtain the equations of equilibrium in the form
E dZ Ui E d*ui
2(1 + a) dx k * 2(1 + ct)(1 - 2a) dx t dx t
These equations can be conveniently rewritten in vector notation. The
quantities dhii/dxjc 2 are components of the vector A u > and dui/dxi = div u.
Thus the equations of equilibrium become
1 J jm 2(1 + a)
AU + -T- grad div u = -p g . (7.2)
1 — La h.
It is sometimes useful to transform this equation by using the vector identity
grad div u = A u+ curl curl u. Then (7.2) becomes
1-2(7
grad div u curl curl u
2(1 -a)
(l + c,)(l-2a)
= " PS E(l-«) ■ (7 ' 3)
We have written the equations of equilibrium for a uniform gravitational
field, since this is the body force most usually encountered in the theory of
elasticity. If there are other body forces, the vector pg on the right-hand
side of the equation must be replaced accordingly.
A very important case is that where the deformation of the body is caused,
not by body forces, but by forces applied to its surface. The equation of
equilibrium then becomes
(1 - 2a) A u + grad div u = (7.4)
or
2(1 - a) grad div u- (1 - 2a) curl curl u = 0. (7.5)
The external forces appear in the solution only through the boundary con-
ditions.
Taking the divergence of equation (7.4) and using the identity
div grad = A>
we find
A div u = 0, (7.6)
i.e. div u (which determines the volume change due to the deformation) is a
harmonic function. Taking the Laplacian of equation (7.4), we then obtain
AAu = 0, (7.7)
i.e. in equilibrium the displacement vector satisfies the biharmonic equation.
These results remain valid in a uniform gravitational field (since the right-
hand side of equation (7.2) gives zero on differentiation), but not in the
general case of external forces which vary through the body.
§7 The equations of equilibrium for isotropic bodies 19
The fact that the displacement vector satisfies the biharmonic equation
does not, of course, mean that the general integral of the equations of equili-
brium (in the absence of body forces) is an arbitrary biharmonic vector; it
must be remembered that the function u(x, y, z) also satisfies the lower-
order differential equation (7.4). It is possible, however, to express the general
integral of the equations of equilibrium in terms of the derivatives of an
arbitrary biharmonic vector (see Problem 10).
If the body is non-uniformly heated, an additional term appears in the
equation of equilibrium. The stress tensor must include the term
-Ka(T-T )S ik
(see (6.2)), and daijcjdxjc accordingly contains a term
-KoidT/dxi^ -[E*l3(l-2<j)]dTldxi.
The equation of equilibrium thus takes the form
— - grad div u curl curl u = a grad T. (7.8)
1 + a 2(1 + a)
Let us consider the particular case of a plane deformation, in which one
component of the displacement vector (u z ) is zero throughout the body,
while the components u x , u y depend only on x and y. The components
u ZZ) Uxz, u yz of the strain tensor then vanish identically, and therefore so do
the components cr xz , o yz of the stress tensor (but not the longitudinal stress
<T ZZ , the existence of which is implied by the constancy of the length of the
body in the jsr-direction). f
Since all quantities are independent of the co-ordinate z, the equations of
equilibrium (in the absence of external body forces) doikjdxjc = reduce in
this case to two equations :
do X z daxy _ dvyx doyy _ ,_ ~.
dx dy dx dy
The most general functions a xx , a xy , a yy satisfying these equations are of
the form
°zx = &x\ty\ °xy = -&xjdxdy, a yy = d 2 x/dx 2 , (7.10)
where x is an arbitrary function of x and y. It is easy to obtain an equation
which must be satisfied by this function. Such an equation must exist, since the
three quantities a xx , a xy , a yy can be expressed in terms of the two quantities
u x , u y , and are therefore not independent. Using formulae (5.13), we find,
for a plane deformation,
o xx + o yy = E(u xx + U yy )l(l + cr)(l - 2a).
t The use of the theory of functions of a complex variable provides very powerful methods of
solving plane problems in the theory of elasticity. See N. I. Muskhelishvili, Some Basic Problems
of the Mathematical Theory of Elasticity, 2nd English ed., P. Noordhoff, Groningen 1963.
20 Fundamental Equations §7
But
du x duy
Vxx+Vyy = AX> U xx + Uyy = — 1 = dlVU,
ox By
and, since by (7.6) div u is harmonic, we conclude that the function x satisfies
the equation
A Ax = 0, (7.11)
i.e. it is biharmonic. This function is called the stress function. When the
plane problem has been solved and the function x is known, the longitudinal
stress o zz is determined at once from the formula
<*zz = oE(u xx + u yy )l(l + cr)(\-2a) = a(a xx + a yy ),
or
<y zz =°Ax- (7.12)
PROBLEMS
Problem 1. Determine the deformation of a long rod (of length /) standing vertically in a
gravitational field.
Solution. We take the 2-axis along the axis of the rod, and the xy-plane in the plane of
its lower end. The equations of equilibrium are doxi/dxt = doytjdxi = 0, da zi /dx { = pg.
On the sides of the rod all the components a t u except a zz must vanish, and on the upper
end (z = I) a xz = a yz = a lz = 0. The solution of the equations of equilibrium satisfying
these conditions is a zz = — pg(l-z), with all other am zero. From one we find «<» to be
u X x = u yy = opg(l—z)jE, u zz = —pg(l—z)/E, u X y — u xt = u yt = 0, and hence by inte-
gration we have the components of the displacement vector, u x — opg(l—z)x/E, Uy ==
o P g(l-z)y/E, u z = -(pg/2E){l 2 -(l-z) 2 -o(x 2 +y*)}. The expression for u z satisfies the
boundary condition u z = only at one point on the lower end of the rod. Hence the solution
obtained is not valid near the lower end.
Problem 2. Determine the deformation of a hollow sphere (of external and internal radii
R 2 and i? 2 ) with a pressure p x inside and p 2 outside.
Solution. We use spherical co-ordinates, with the origin at the centre of the sphere.
The displacement vector u is everywhere radial, and is a function of r alone. Hence curl u=0,
and equation (7.5) becomes grad div u = 0. Hence
1 d(r*u)
div u = — = constant = 3a,
r 2 dr
or u = ar+bjr*. The components of the strain tensor are (see formulae (1.7)) u„ == a—2b/r 3 ,
uee = «00 = a+b/r*. The radial stress is
E E 2E b
&rr = — — —A{\ — a)Urr + 2aU ee } = —a
(l + cr)(l-2(7) lv ' w l-2a l + o-r3
The constants a and b are determined from the boundary conditions: a„ = — p x at r — R u
and o„ — —pi at r — i?g. Hence we find
- P1 R 1*-P2 R # 1 ~ 2a h _ Rl 3R 2 3 (pl-p2) 1 + ct
R 2 3 -Ri 3 ' E ' " i^-fli 3 '~2E'
§7 The equations of equilibrium for isotropic bodies 21
For example, the stress distribution in a spherical shell with a pressure pi=* p inside and
p 2 = outside is given by
pR?l Ra 3 \ pR$ I Rs? \
For a thin spherical shell of thickness h — R 2 —Rj, <^ R we have approximately
u = pR 2 (l - a)l2Eh, a ee = o H = IpRjh, o rr = \p,
where a„ is the mean value of the radial stress over the thickness of the shell.
The stress distribution in an infinite elastic medium with a spherical cavity (of radius R)
subjected to hydrostatic compression is obtained by putting i? x = R, i^ = oo, p x = 0,
Ps = p:
/i RZ \ I R3 \
°rr = -P\\ ~ — j, o„ = o„ = -p\\+—y
At the surface of the cavity the tangential stresses a ee = a^ = —3p/2, i.e. they exceed the
pressure at infinity.
Problem 3. Determine the deformation of a solid sphere (of radius R) in its own gravi-
tational field.
Solution. The force of gravity on unit mass in a spherical body is —gt/R. Substituting
this expression in place of g in equation (7.3), we obtain the following equation for the radial
displacement:
E{\-a) d/ld(r2«)
- a) d / 1 d(r%) \ r
-2a)M72 dr / = P8 R'
(l + <r)(l-2a)
The solution finite for r — which satisfies the condition a rr = for r — R is
#>fl(l-2a)(l + <r) /3-a r2
u = r
105(1 -a)
IS- a r* \
It should be noticed that the substance is compressed (u rr < 0) inside a spherical surface of
radius i?\/{(3— <0/3(l +a)} and stretched outside it (u rr > 0). The pressure at the centre of
the sphere is (3 —a)g pRj 10(1 —o).
Problem 4. Determine the deformation of a cylindrical pipe (of external and internal radii
jR 8 and i?i), with a pressure p inside and no pressure outside.f
Solution. We use cylindrical co-ordinates, with the ar-axis along the axis of the pipe.
When the pressure is uniform along the pipe, the deformation is a purely radial displacement
w r = u(r). Similarly to Problem 2, we have
1 d(m)
div u = — = constant = 2a.
r dr
Hence u = ar+bjr. The non-zero components of the strain tensor are (see formulae (1.8))
Mr, = dtt/dr = a—b/r 2 , u^ = u/r = a+b/r*. From the conditions a„ = at r = R t ,
and o„ = —p at r = R lt we find
pR? (l + g)(l-2or) pR^RJ l + o-
a ~R 2 2 -Ri*' E ' ~ R 2 *-R!* '~E~'
t In Problems 4, 5 and 7 it is assumed that the length of the cylinder is maintained constant, so
that there is no longitudinal deformation.
22 Fundamental Equations §7
The stress distribution is given by the formulae
a zz = 2paR^/(R 2 z -Ri z ).
Problem 5. Determine the deformation of a cylinder rotating uniformly about its axis.
Solution. Replacing the gravitational force in (7.3) by the centrifugal force pftV (where
ft is the angular velocity), we have in cylindrical co-ordinates the following equation for the
displacement u f = u(r):
? (l- g ) £/ljM\ _
a)(l-2a)dr\r dr J H
E(l-a) d/1 d(ru)
(1 + o-)(1-2(t) dr\r dr ,
The solution which is finite for r = and satisfies the condition a„ = for r = R is
" = —sm^r r[{)l
Problem 6. Determine the deformation of a non-uniformly heated sphere with a spherically
symmetrical temperature distribution.
Solution. In spherical co-ordinates, equation (7.8) for a purely radial deformation is
d/1 d(r^u)\ l + o- dr
dr\r 2 dr / 3(1 -a) dr
The solution which is finite for r = and satisfies the condition a„ — for r = R is
v '
The temperature T(r) is measured from the value for which the sphere, if uniformly heated,
is regarded as undeformed. In the above formula the temperature in question is taken as that
of the outer surface of the sphere, so that T(R) = 0.
Problem 7. The same as Problem 6, but for a non-uniformly heated cylinder with an
axially symmetrical temperature distribution.
Solution. We similarly have in cylindrical co-ordinates
v ' o
Problem 8. Determine the deformation of an infinite elastic medium with a given tempera-
ture distribution T(x, y, z) which is such that the temperature tends to a constant value To
at infinity, there being no deformation there.
Solution. Equation (7.8) has an obvious solution for which curl u = and
div u = <x(l + o)[T(x, y, z) - T J/3(1 - a).
§7 The equations of equilibrium for isotropic bodies 23
The vector u, whose divergence is a given function defined in all space and vanishing at
infinity, and whose curl is zero identically, can be written, as we know from vector analysis,
in the form
u(x, y, z) = - — grad dV ,
4n J r
where
r = ^{( X - X >)2 + (y-y') 2 + (z-z') 2 }.
We therefore obtain the general solution of the problem in the form
a(l+a) ^CT'-Tq
u = —
127T(1-Cr)
-a) CI -To
_jL grad ° dV ', (1)
— cr) J r
wHere T' == T(x', y\ z').
If a finite quantity of heat q is evolved in a very small volume at the origin, the temperature
distribution can be written T—T = (qlC)S(x)S(y)8(z), where C is the specific heat of the
medium. The integral in (1) is then qlCr, and the deformation is given by
a(l + a)q T
u =
12tt(1 -a)C r3
Problem 9. Derive the equations of equilibrium for an isotropic body (in the absence of
body forces) in terms of the components of the stress tensor.
Solution. The required system of equations contains the three equations
daaddXk = (1)
and also the equations resulting from the fact that the six different components of k<* are
not independent quantities. To derive these equations, we first write down the system of
differential relations satisfied by the components of the tensor «<*. It is easy to see that the
quantities
1 / dUi dujc\
2\dxjc dxil
satisfy identii ally the relations
d 2 Uik d 2 ui m d 2 uu d 2 uje T
•+ . . = — — + ■
dxidx m dxfdxjc dxjcdx m dx%dxi
Here there are only six essentially different relations, namely those corresponding to the fol-
lowing values of i, k, I, m : 1122, 1133,2233, 1123, 2213,3312. All these are retained if the above
tensor equation is contracted with respect to / and m :
d 2 U tt _ d 2 Uu d 2 Ukl
dxidxjc dxjcdxi dxidxi
Substituting here unc in terms of o,a: according to (5.12) and using (1), we obtain the re-
quired equations:
(l + o)AcTi k +—^- = 0. (3)
OXfOXjc
These equations remain valid in the presence of external forces constant throughout the body.
24 Fundamental Equations §7
Contracting equation (3) with respect to the suffixes i and k, we find that A<*ii = 0, i.e.
<tji is a harmonic function. Taking the Laplacian of equation (3), we then find that AA"<* = 0,
i.e. the components ant are biharmonic functions. These results follow also from (7.6)
and (7.7), since aw and unc are linearly related.
Problem 10. Express the general integral of the equations of equilibrium (in the absence
of body forces) in terms of an arbitrary biharmonic vector (B. G. Galerkin 1930).
Solution. It is natural to seek a solution of equation (7.4) in the form
u = Af +A grad div f.
Hence div u = (1 +A) div Af. Substituting in (7.4), we obtain
(l-2a)AAf+[2(l-(r)^ + l] grad div Af= 0.
From this we see that, if f is an arbitrary biharmonic vector (A Af = 0)> then
u = Af - — r grad divf.
2(1 - a)
Problem 1 1 . Express the stresses a„, a^, a r $ for a plane deformation (in polar co-ordinates
r, <j>) as derivatives of the stress function.
Solution. Since the required expressions cannot depend on the choice of the initial line
of <f>, they do not contain <f> explicitly. Hence we can proceed as follows : we transform the
Cartesian derivatives (7.10) into derivatives with respect to r, <f>, and use the results that
a„ = (axs^.o, o H = (a tfy )^_ , <^ = (<^i/)0=o, the angle <f> being measured from the *-axis.
Thus
arr = ~rJr + ^ W ° H ~ dr2 ' ^ ~ dr \ r d V
Problem 12. Determine the stress distribution in an infinite elastic medium containing
a spherical cavity and subjected to a homogeneous deformation at infinity.
Solution. A general homogeneous deformation can be represented as a combination of a
homogeneous hydrostatic extension (or compression) and a homogeneous shear. The former
has been considered in Problem 2, so that we need only consider a homogeneous shear.
Let <7ij; (0) be the homogeneous stress field which would be found in all space if the cavity
were absent: in a pure shear a«<°> = 0. The corresponding displacement vector is denoted
by u<°>, and we seek the required solution in the form u = uC'+u' 1 ', where the function u<*>
arising from the presence of the cavity is zero at infinity.
Any solution of the biharmonic equation can be written as a linear combination of centrally
symmetrical solutions and their spatial derivatives of various orders. The functions r\
r, 1, 1/r are independent centrally symmetrical solutions. Hence the most general form of a
biharmonic vector u (1> , depending only on the components of the constant tensor ane w
as parameters and vanishing at infinity, is
Ui a) = A(m «»—[-)+Bo kl «» - )+Ca k P r. (1)
dx k \r) dxidx k dxi\rj dxidx k dxi
Substituting this expression in equation (7.4), we obtain
{1 _2a)^+4-^ L = [2(l-2cr)C+(A + 2C)]o kl « » l\ I = 0,
v dx? dx t dxi dxidx k dxi r
whence A = — 4C(1 — a). Two further relations between the constants A, B, C are obtained
from the condition at the surface of the cavity: (<7,-fc< > + <r<*: (1) )rt* = for r = R (R being the
§8 Equilibrium of an elastic medium bounded by a plane 25
radius of the cavity, the origin at its centre, and n a unit vector parallel to r). A somewhat
lengthy calculation, using (1), gives the following values:
B = Ci? 2 /5, C = 5#3(1 + a )j2E(7 - 5 a).
The final expression for the stress distribution is
5(l-2a)/ J R\s 3 jr^
I 5(\-2a)/R\ A 3 (R\*\
l 1+J ^r(7) + 7^l7) ) +
15 /R\ 3 ( /R\ 2 \
+ —-—( — ) CT -(~ ) \(oii (0) nkni+o k Pnini) +
15 /R\*l /R\ 2 \ , ns
+ 2(7^)(T) 3 ( 1 - 2CT -(7) 2 | 8 " < ' iAB -
In order to obtain the stress distribution for arbitrary one*- ) (not a pure shear), ao: (0) in
this expression must be replaced by ct,* (0) — £§<& ff» (0) , and the expression
r R3 l
corresponding to a deformation homogeneous at infinity (cf. Problem 2) must be added. We
may give here the general formula for the stresses at the surface of the cavity:
15 l
a ik = - — — {(1 - (y){(Tik i0) - crtPmnjc- a k i i0) nini) +
7 — 5cr l
5CT-1 )
+ vim m nin m nin k - crcW *^^ * + — — <Jii {0) (8ik - nm) \.
Near the cavity, the stresses considerably exceed the stresses at infinity, but this extends
over only a short distance (the concentration of stresses). For example, if the medium is
subjected to a homogeneous extension (only o lz W different from zero), the greatest stress
occurs on the equator of the cavity, where
27- 15a
§8. Equilibrium of an elastic medium bounded by a plane
Let us consider an elastic medium occupying a half-space, i.e. bounded
on one side by an infinite plane, and determine the deformation of the
26 Fundamental Equations §8
medium caused by forces applied to its free surf ace. f The distribution of
these forces need satisfy only one condition: they must vanish at infinity in
such a way that there is no deformation at infinity. In such a case the equa-
tions of equilibrium can be integrated in a general form.
The equation of equilibrium (7.4) holds throughout the space occupied
by the medium:
graddivu + (l-2cr)Au = 0. (8.1)
We seek a solution of this equation in the form
u = f+grad<£, (8.2)
where <£ is some scalar and the vector f satisfies Laplace's equation:
Af=0. (8.3)
Substituting (8.2) in (8.1), we then obtain the following equation for <f>:
2(l-a)A<£ = -divf. (8.4)
We take the free surface of the elastic medium as the #y-plane; the medium
is in z > 0. We write the functions f x and f y as the ^-derivatives of some
functions g x and g y :
fx = dgxfiz* fy = dgy/dz- (8-5)
Since f x and f y are harmonic functions, we can always choose the functions
g x and g y so as to satisfy Laplace's equation :
Ag* = 0, Agy = 0. (8.6)
Equation (8.4) then becomes
2(1-.)A^= --(— +—+/,).
Since g x , gy and f z are harmonic functions, we easily see that a function
<j> which satisfies this equation can be written as
where i/j is again a harmonic function:
A«A = o. ( 8 - 8 )
| The most direct and regular method of solving this problem is to use Fourier's method on
equation (8.1). In that case, however, some fairly complicated integrals have to be calculated. The
method given below is based on a number of artificial devices, but the calculations are simpler.
z-0
= -2(1 + a)P z [E. (8.10)
z =
§8 Equilibrium of an elastic medium bounded by a plane 27
Thus the problem of determining the displacement u reduces to that
of finding the functions g Xy g yy f z , iff, all of which satisfy Laplace's equation.
We shall now write out the boundary conditions which must be satisfied at
the free surface of the medium (the plane z = 0). Since the unit outward
normal vector n is in the negative ^-direction, it follows from the general
formula (2.8) that a iz = -P t . Using for a ik the general expression (5.11) and
expressing the components of the vector u in terms of the auxiliary quantities
g x> g y , f z and «/r, we obtain after a simple calculation the boundary conditions
r d 2 gx -\ Vdl l-2ar _ 1 /dgz [ dg v \ +2 #j-
L 5* 2 J z« o + ldx[ 2(1 - of* 2(1 - a) \ dx dy) dz).
= -2(1 +*)P X /E, (8.9)
La^ 2 Jz-o Uvl2(l-ar 2(l-ff)U* 3y/ aWJz=o
= -2(l + a)P 2 ,/£ , ,
[K'-(t4') rf 3
The components P x , P y , P z of the external forces applied to the surface are
given functions of the co-ordinates x and y, and vanish at infinity.
The formulae by which the auxiliary quantities g X} g y ,f z and \jj were defined
do not determine them uniquely. We can therefore impose an arbitrary
additional condition on these quantities, and it is convenient to make the
quantity in the braces in equations (8.9) vanish :f
dy
Then the conditions (8.9) become simply
raagfc-i _ 2(l + o) ra^-l _ 2(l + a) p
La^Lo * *' L"^"J z =o~ £ ^
Equations (8.l0)-(8.l2) suffice to determine completely the harmonic
functions g x , g y , f z and ifj.
For simplicity, we shall consider the case where the free surface of an
elastic half-space is subjected to a concentrated force F, i.e. one which is
applied to an area so small that it can be regarded as a point. The effect of
this force is the same as that of surface forces given by P = FS(x)8(y), the
origin being at the point of application of the force. If we know the solution
for a concentrated force, we can immediately find the solution for any force
distribution P(#, y). For, if
Ui = G ik (x,y, z)F k (8.13)
f We shall not prove here that this condition can in fact be imposed; this follows from the absence
of contradiction in the result.
( i - 2 ^-(S + l ! ) +4(1 - <7) S= o - (8 - n)
(8.12)
28 Fundamental Equations §8
is the displacement due to the action of a concentrated force F applied at
the origin, then the displacement caused by forces T*{x,y) is given by the
integralf
Ui = f J G«(*-*\ y-y\ z)P k (x', y') dx' d/. (8.14)
We know from potential theory that a harmonic function / which is zero
at infinity and has a given normal derivative df/dz on the plane z = is
given by the formula
1 rr\df(x\y\z)l dx'dy'
f(x, y, z) = ,
2ttJ J L dz Jz = o r
where
r = \/{(x — x') 2 + (y—y') 2 + z 2 }.
Since the quantities dg x /dz, dgyjdz and that in the braces in equation (8.10)
satisfy Laplace's equation, while equations (8.10) and (8.12) determine the
values of their normal derivatives on the plane z = 0, we have
/,_ &t + ^L\ J± . i±f f [**<>?> u d/
\ 8* 9y / dz ttE J J r
\ + a F t
tE r
dgx
l + o F x
dgy 1+ff Fy
dz
ttE ' r'
dz ttE r'
(8.15)
(8.16)
r oz ttCj r
where now r = ^(x 2 +y 2 + z 2 ).
The expressions for the components of the required vector u involve the
derivatives of g Xi g y with respect to x, y, z, but not g x , g y themselves. To
calculate dg x jdx, dgyjdy, we differentiate equations (8.16) with respect to
x and y respectively :
d 2 g x _ _l + a F^ d 2 gy = 1 + or F y y
dxdz ttE r 3 ' dydz ttE r 3
Now, integrating over z from oo to z, we obtain
dg x 1+ a F x x
dx ttE ' r(r + z)'
dgy 1 + a F p y
(8.17)
dy ttE r{r + z)
We shall not pause to complete the remaining calculations, which are
elementary but laborious. We determine f z and diff/dz from equations (8.11),
t In mathematical terms, Gw is the Green's tensor for the equations of equilibrium of a semi-infinite
medium.
§8
Equilibrium of an elastic medium bounded by a plane
29
(8.15) and (8.17). Knowing di/jjdz, it is easy to calculate dtfjjdx and difjjdy by
integrating with respect to z and then differentiating with respect to x and y.
We thus obtain all the quantities needed to calculate the displacement vector
from (8.2), (8.5) and (8.7). The following are the final formulae:
l + <r(
xz (l-2,r)*-| 2(l-a)r+z
—\r z -\ v x -r
r 3 r(r + z) J r(r + z)
\2r(ar + z) + z 2 ]x )
r 6 {r + z)* )
Uy _ Ltfff^ - (1-2^1 2(1- CT ), + , I
+
(r + #) J " r(r + z)
[2r(ar + z) + z 2 ]y
r 3 (r + z) 2
(xF x +yF y )\,
1 + erf r2(l — CT ) z 2 i r 1-2ct zi )
In particular, the displacement of points on the surface of the medium is
given by putting z = 0:
1 + cr 1
u x
Uu =
(l-2o>
: lax \
-F z + 2(l-a)F x +-—(xF x +yFy)\,
2ttE r\r r 2 )
1 + alf (1-2ct)v lay \
— -.- - -^ ^F z + 2(l-a)F y + -/(xF x +yF y )\,
2ttE r{ r r 2 j
(8.19)
»* = "T^-- k 1 " "X^ + O ~ 2ct ) " («^r+^y)
PROBLEM
Determine the deformation of an infinite elastic medium when a force F is applied to a
small region in it. t
Solution. If we consider the deformation at distances r which are large compared with
the dimension of the region where the force is applied, we can suppose that the force is
applied at a point. The equation of equilibrium is (cf. (7.2))
Au +
1
l-2o-
grad div u =
2(1 + a)
E
Fo(r),
(1)
where 8(r) = S(x)8(y)8(z), the origin being at the point where the force is applied. We seek
the solution in the form u = Uo+Ux, where u„ satisfies the Poisson-type equation
A«o = - 2(1 J~ g) Fo(r).
E
(2)
■f The corresponding problem for an arbitrary infinite anisotropic medium has been solved by
I. M. Lifshitz and L. N. RozentsveIg (Zhttrnal experimental' not i teoreticheskoi fiziki 17, 783, 1947).
30 Fundamental Equations §9
We then have for ui the equation
grad div ui + (l — 2o)/\vl\ = — grad div un. (3)
The solution of equation (2) which vanishes at infinity is uo = (1 + o)F/2nEr. Taking the
curl of equation (3), we have A curl ui = 0. At infinity we must have curl ui = 0. But a
function harmonic in all space and zero at infinity must be zero identically. Thus curl ui = 0,
and we can therefore write ui = grad^. From (3) we obtain grad {2(1 — a) A^ + div uo} = 0.
Hence it follows that the quantity in braces is a constant, and it must be zero at infinity; we
therefore have in all space
dlVUo 1 + ct /1\
£</, = = F- grad (-).
r 2(1 -a) 4ttE(1-ct) & \rf
If i/i is a solution of the equation A«A = l/ r > then
1 + CT
6 = F« grad ib.
Y 4ttE(1-o)
Taking the solution iff = \r, which has no singularities, we obtain
1 + ct (F.n)n-F
ui = grad </> = — — ,
57r.c(l — ct) r
where n is a unit vector parallel to the radius vector r. The final result is
1 + CT (3-4<r)F+n(n.F)
$7rE(l-a) r
On putting this formula into the form (8.13) we obtain the Green's tensor for the equa-
tions of equilibrium of an infinite isotropic medium:!
1 + CT 1
87tE(1 — ct) r
1
Hk
1 d*r
47Tfil r 4(1 — ct) dxfdxjc
§9. Solid bodies in contact
Let two solid bodies be in contact at a point which is not a singular point
on either surface. Fig. la shows a cross-section of the two surfaces near
the point of contact O. The surfaces have a common tangent plane at O,
which we take as the xy-plane. We regard the positive ^-direction as being
into either body (i.e. in opposite directions for the two bodies) and denote
the corresponding co-ordinates by z and z'.
f The fact that the components of the tensor Gxk are first-order homogeneous functions of the co-
ordinates x, y, z is evident from arguments of homogeneity applied to the form of equation (1), where
the left-hand side is a linear combination of the second derivatives of the components of the vector u,
and the right-hand side is a third-order homogeneous function (S(ar) = a _3 S(r)).
This property remains valid in the general case of an arbitrary anisotropic medium.
§9
Solid bodies in contact
31
Near a point of ordinary contact with the acy-plane, the equation of the
surface can be written
z = K afi x^ fii (9.1)
where summation is understood over the values 1, 2, of the repeated suffixes
a, j8 (xi = x, x 2 = y), and K afi is a symmetrical tensor of rank two, which
characterises the curvature of the surface: the principal values of the tensor
K ap are l/2Ri and 1/2^2, where Ri and R 2 are the principal radii of curvature
of the surface at the point of contact. A similar relation for the surface of
the other body near the point of contact can be written
Z' = K'apXJCfi. (9-2)
Let us now assume that the two bodies are pressed together by applied
forces, and approach a short distance A.f Then a deformation occurs near
the original point of contact, and the two bodies will be in contact over a
small but finite portion of their surfaces. Let u z and u' z be the components
(along the * and z' axes respectively) of the corresponding displacement
vectors for points on the surfaces of the two bodies. The broken lines
Fig. 1
in Fig. lb show the surfaces as they would be in the absence of any deforma-
tion, while the continuous lines show the surfaces of the deformed bodies; the
letters z and z' denote the distances given by equations (9.1) and (9.2). It
is seen at once from the figure that the equation
(z + u z ) + (z' + u' z ) = h,
or
(*«*+*'«*)*«**+«« + «'« = n >
(9.3)
f This contact problem in the theory of elasticity was first solved by H. Hertz.
32 Fundamental Equations §9
holds everywhere in the region of contact. At points outside the region of
contact, we have
z + z' + u z +u' z < h.
We choose the x andj> axes to be the principal axes of the tensor K afi + K afi .
Denoting the principal values of this tensor by A and B,-\ we can rewrite
equation (9.3) as
Ax* + By 2 + u z + u' z = h. (9.4)
We denote by P z {x, y) the pressure between the two deformed bodies at
points in the region of contact; outside this region, of course P z = 0. To
determine the relation between P z and the displacements u z> u' z , we can
with sufficient accuracy regard the surfaces as plane and use the formulae
obtained in §8. According to the third of formulae (8.19) and (8.14), the
displacement u z under the action of normal forces P z (x, y) is given by
~ ttE J J"
u z = -^-\\-^-dx'dy,
Mz = -^-JJ-T— dxdy >
where a, a' and E, E' are the Poisson's ratios and the Young's moduli of the
two bodies. Since P z = outside the region of contact, the integration ex-
tends only over this region. It may be noted that, from these formulae, the
ratio u z \u' z is constant:
u z \u' z = (1 - CT 2)£'/(1 - CT '2)£. (9.6)
The relations (9.4) and (9.6) together give the displacements u Zi u' z at every
point of the region of contact (although (9.5) and (9.6), of course, relate to
points outside that region also).
Substituting the expressions (9.5) in (9.4), we obtain
l/l-o* l-<x'2\ r rP z (x',y')
-I— — + — jp— ) j j r dx' d/ = h-Ax*-By2. (9.7)
t The quantities A and B are relaced to the radii of curvature i? 1( R 2 and R\, R' t by
2(A + B) = — +— +—+—,
V ' R! R 2 R\ R' 2
/ 1 1 \ 2 / 1 1 \2
T \Ri i? 2 /U'i rJ'
where <f> is the angle between the normal sections whose radii of curvature are R t and R\.
The radii of curvature are regarded as positive if the centre of curvature lies within the body con-
cerned, and negative in the contrary case.
§9 Solid bodies in contact 33
This integral equation determines the distribution of the pressure P z over
the region of contact. Its solution can be found by analogy with the following
results of potential theory. The idea of using this analogy arises as follows:
firstly, the integral on the left-hand side of equation (9.7) is of a type com-
monly found in potential theory, where such integrals give the potential of a
charge distribution; secondly, the potential inside a uniformly charged
ellipsoid is a quadratic function of the co-ordinates.
If the ellipsoid x 2 la 2 +y 2 jb 2 + z 2 lc 2 = 1 is uniformly charged (with volume
charge density p), the potential in the ellipsoid is given by
<K x > y> z )
00
n x 2 y 2 z 2 \ d£
= irpabcj |1 - — - -^j - J^r] V{{a 2 + i){b 2 + £ ){c 2 + t )} '
o
In the limiting case of an ellipsoid which is very much flattened in the
z- direction (c -> 0), we have
00
Hx,y) = «pabcj{l -JL-- _^j ;
in passing to the limit c -> we must, of course, put z = for points inside
the ellipsoid. The potential <j>{x, y, z) can also be written as
,_ c c c p dx ' d y' dz '
mytZ) ~ J J J v{(^-^) 2 +(^-y) 2 +(^-^) 2 } ,
where the integration is over the volume of the ellipsoid. In passing to the
limit c ->0, we must put z = z' = in the radicand; integrating over z'
between the limits
±c^{\-(x' 2 \a 2 )-{y' 2 \b 2 )\
we obtain
r rdx' dy' 1/ x' 2 y' 2 \
where
r= V{{x-x') 2 + {y-y'Y},
and the integration is over the area inside the ellipse
x '2j a 2 +y '2j b 2 = 1.
34 Fundamental Equations §9
Equating the two expressions for <f>(x, v), we obtain the identity
r rdx' dy' If x' 2 y' 2 \
J J ~~r V \ ~a~ 2 W)
Comparing this relation with equation (9.7), we see that the right-hand
sides are quadratic functions of x and y of the same form, and the left-hand
sides are integrals of the same form. We can therefore deduce immediately
that the region of contact (i.e. the region of integration in (9.7)) is bounded
by an ellipse of the form
x 2 y 2
and that the function P z (x, y) must be of the form
Pfay) = constant x ^ (l - *- - £).
Taking the constant such that the integral jjP z dx dy over the region of
contact is equal to the given total force F which moves the bodies together,
we obtain
'^-zWf 1 -*-£)• (9 ' 10)
This formula gives the distribution of pressure over the area of the region of
contact. It may be pointed out that the pressure at the centre of this region
is f times the mean pressure FjTrab.
Substituting (9.10) in equation (9.7) and replacing the resulting integral
in accordance with (9.8), we obtain
00
FDr/ x 2 y 2 \
-ft 1 -*rr£i) "fMc+flc+flo
= h-Ax 2 -By\
where
4\ E + E' )'
This equation must hold identically for all values of x and y inside the
§9 Solid bodies in contact 35
ellipse (9.9) ; the coefficients of x and y and the free terms must therefore be
respectively equal on each side. Hence we find
FDr d|
1 g (9.11)
tu r
TT J
V{(<* 2 + Z)(b 2 + m'
FDr dg
bD r
A = — -
77 J (d
B
FDr d£
bu r
IT J I
(b 2 + i)V{(a 2 + ZW + m
Equations (9.12) determine the semi-axes a and b of the region of contact
from the given force F {A and B being known for given bodies). The
relation (9.11) then gives the distance of approach h as a function of the force
F. The right-hand sides of these equations involve elliptic integrals.
Thus the problem of bodies in contact can be regarded as completely
solved. The form of the surfaces (i.e. the displacements u z , u' z ) outside the
region of contact is determined by the same formulae (9.5) and (9.10); the
values of the integrals can be found immediately from the analogy with the
potential outside a charged ellipsoid. Finally, the formulae of §8 enable us to
find also the deformation at various points in the bodies (but only, of course,
at distances small compared with the dimensions of the bodies).
Let us apply these formulae to the case of contact between two spheres of
radii R and R'. Here A = B = 1/2R+1/2R'. It is clear from symmetry
that a = b, i.e. the region of contact is a circle. From (9.12) we find the
radius a of this circle to be
a = Fii3{DRR'l(R + R')}V3. (9.13)
h is in this case the difference between the sum R + R' and the distance be-
tween the centres of the spheres. From (9.10) we obtain the following
relation between F and h :
h _ F^Jz^l+l)] 1 ' 3 . (9.14)
It should be noticed that h is proportional to F m ; conversely, the force F
varies as h 3/2 . We can write down also the potential energy U of the spheres
in contact. Since —F= —dU/dh, we have
2 / RR'
U = #5/2 —
SD
I RR'
JrTW < 9 - 15 >
Finally, it may be mentioned that a relation of the form h = constant x F 2/3 ,
or F = constant x A 3/2 , holds not only for spheres but also for other finite
36 Fundamental Equations §9
bodies in contact. This is easily seen from similarity arguments. If we make
the substitution
a 2 -> oca 2 , b 2 -> <xb 2 , F -> a 3/2 F,
where a is an arbitrary constant, equations (9.12) remain unchanged. In
equation (9.11), the right-hand side is multiplied by a, and so h must be
replaced by aJi if this equation is to remain unchanged. Hence it follows
that F must be proportional to h 3/2 .
PROBLEMS
Problem 1. Determine the time for which two colliding elastic spheres remain in contact.
Solution. In a system of co-ordinates in which the centre of mass of the two spheres is
at rest, the energy before the collision is equal to the kinetic energy of the relative motion
£/«>*> where v is the relative velocity of the colliding spheres and /x = mjW^/Cwii+wia) their
reduced mass. During the collision, the total energy is the sum of the kinetic energy, which
may be written £/Jt 8 , and the potential energy (9.15). By the law of conservation of energy
we have
(dh\ 2 .„,„ , 4 / RR'
lahY 4 /
J— +kh 5/2 = u.v 2 , k = — A
r \dt) r 5ZW-
R + R'
The maximum approach h of the spheres corresponds to the time when their relative velocity
h = 0, and is h = {nlk)*'W*.
The time t during which the collision takes place (i.e. h varies from to h and back) is
h„ i
dx
T
Jv(» 2 -^ /2 //x) Uv J
V(*> 2 - Mfi'*/p) \ k 2 v] J V(i - * 2/5 ) '
or
51X9/10) IW l#W '
By using the statical formulae obtained in the text to solve this problem, we have neglected
elastic oscillations of the spheres resulting from the collision. If this is legitimate, the velocity
v must be small compared with the velocity of sound. In practice, however, the validity of
the theory is limited by the still more stringent requirement that the resulting deformations
should not exceed the elastic limit of the substance.
Problem 2. Determine the dimensions of the region of contact and the pressure distri-
bution when two cylinders are pressed together along a generator.
Solution. In this case the region of contact is a narrow strip along the length of the
cylinders. Its width 2a and the pressure distribution across it can be found from the formulae
in the text by going to the limit bja -*■ oo. The pressure distribution will be of the form
Pz(x) = constant X \/(l —x 3 la 2 ), where x is the co-ordinate across the strip; normalising
the pressure to give a force F per unit length, we obtain
IF
IF II x*\
Substituting this expression in (9.7) and effecting the integration by means of (9.8), we have
4DF f df 8DF
A =
i
3tt J (a 2 + £) 3/2 | 3-rra 2
§10 The elastic properties of crystals 37
One of the radii of curvature of a cylindrical surface is infinite, and the other is the radius
of the cylinder; in this case, therefore, A = 1/2R+112R', B = 0. We have finally for the
width of the region of contact
16DF RR'
I/16DF RR' \
a= VI 3tt ' R + R')
§10. The elastic properties of crystals
The change in the free energy in isothermal compression of a crystal is, as
with isotropic bodies, a quadratic function of the strain tensor. Unlike what
happens for isotropic bodies, however, this function contains not just two
coefficients, but a larger number of them. The general form of the free energy
of a deformed crystal is
F = ^iklmUikUim, (10.1)
where X iJc i m is a tensor of rank four, called the elastic modulus tensor. Since
the strain tensor is symmetrical, the product uacUim is unchanged when the
suffixes i, k, or /, m, or /, / and k, m, are interchanged. Hence we see that the
tensor Xmm can be defined so that it has the same symmetry properties:
Xiklm = Xmm = ^ikml — hmik> (10.2)
A simple calculation shows that the number of different components of a
tensor of rank four having these symmetry properties is in general 21.
In accordance with the expression (10.1) for the free energy, the stress
tensor for a crystal is given in terms of the strain tensor by
one = dFjdUije — XikimUim't (10.3)
cf. also the last footnote to this section.
If the crystal possesses symmetry, relations exist between the various
components of the tensor X ik i m , so that the number of independent com-
ponents is less than 21.
We shall discuss these relations for each possible type of macroscopic
symmetry of crystals, i.e. for each of the crystal classes, dividing these into the
corresponding crystal systems.
(1) Triclinic system. Triclinic symmetry (classes C\ and C*) does not place
any restrictions on the components of the tensor Xikim, and the system of co-
ordinates may be chosen arbitrarily as regards the symmetry. All the 21
moduli of elasticity are non-zero and independent. However, the arbitrariness
of the choice of co-ordinate system enables us to impose additional conditions
on the components of the tensor Xikim- Since the orientation of the co-ordinate
system relative to the body is defined by three quantities (angles of rotation),
there can be three such conditions; for example, three of the components may
38 Fundamental Equations §10
be taken as zero. Then the independent quantities which describe the elastic
properties of the crystal will be 18 non-zero moduli and 3 angles defining the
orientation of the axes in the crystal.
(2) Monoclinic system. Let us consider the class C s ; we take a co-ordinate
system with the *y-plane as the plane of symmetry. On reflection in this
plane, the co-ordinates undergo the transformation x -> x, y -+y, z -> — z.
The components of a tensor are transformed as the products of the corres-
ponding co-ordinates. It is therefore clear that, in the transformation men-
tioned, all components X ik i m whose suffixes include z an odd number of
times (1 or 3) will change sign, while the other components will remain un-
changed. By the symmetry of the crystal, however, all quantities characterising
its properties (including all components Xiicim) must remain unchanged on
reflection in the plane of symmetry. Hence it is evident that all components
with an odd number of suffixes z must be zero. Accordingly, the general
expression for the elastic free energy of a crystal belonging to the monoclinic
system is
* — iX xxxx u xx +2^yyyy u yy + 2^zzzzUzz 2 + XxxyyUxxUyy + X X x ZZ UxxUzz +
+ AyyzzUyyU zz + 2X X yxyU X y 2 + 2X xzxz U xz 2 + 2Xy zyz Uy z 2 + 2\ X xxyUxxU xy +
+ 2Xyyy x UyyUy x + 2Xxy ZZ U X yU zz + 4X ZZ yzUx Z Uy z . ( 1 0.4)
This contains 13 independent coefficients. A similar expression is obtained
for the class C2, and also for the class C^n, which contains both symmetry
elements (C2 and an). In the argument given, however, the direction of only
one co-ordinate axis (that of z) is fixed ; those of x and y can have arbitrary
directions in the perpendicular plane. This arbitrariness can be used to make
one coefficient, say X xyzz , vanish by a suitable choice of axes. Then the 13
quantities which describe the elastic properties of the crystal will be 12 non-
zero moduli and one angle defining the orientation of the axes in the xy-plane.
(3) Orthorhombic system. In all the classes of this system (Cz v , D 2 , £>2ft) the
choice of co-ordinate axes is determined by the symmetry, and the expression
obtained for the free energy is the same for each class.
Let us consider, for example, the class D^n', we take the three planes of
symmetry as the co-ordinate planes. Reflections in each of these planes are
transformations in which one co-ordinate changes sign and the other two
remain unchanged. It is evident therefore that the only non-zero components
X-ikim are those whose suffixes contain each of x, y, z an even number of times ;
the other components would have to change sign on reflection in some plane
of symmetry. Thus the general expression for the free energy in the ortho-
rhombic system is
F — \XxxxxUxz +2*yyyyUyy +^X zzzz U zz 2 + X X xyyU X xUyy + XxxzzUxxUzz +
+ Xyy ZZ UyyU ZZ + 2X X y x yU X y 2 + 2X XZXZ U XZ % + 2Xy Z y Z Uy z 2 . (10.5)
It contains nine moduli of elasticity.
§10 The elastic properties of crystals 39
(4) Tetragonal system. Let us consider the class C 4t ,; we take the axis C 4
as the #-axis, and the x and y axes perpendicular to two of the vertical
planes of symmetry. Reflections in these two planes signify transformations
x -> — x, y -+y, z ->z
and
x -> x, y -> —y, z-+z;
all components X ik im with an odd number of like suffixes therefore vanish.
Furthermore, a rotation through an angle £tt about the axis C 4 is the trans-
formation
x -+y, v -> —x, z -+z.
Hence we have
Xxxxx = "yyyyy X X xzz — Ayyzz> X X zxz = Ayzyz-
The remaining transformations in the class C iv do not give any further
conditions. Thus the free energy of crystals in the tetragonal system is
F =» %\xxxx(Uxx 2 + Uyy 2 ) + &zzzzUzz 2 + hzxzz(UxzUzz + UyyU zz ) +
+ ^xxyyUxxUyy + 2X X yxyU X y 2 + 1\ X zxz{Uxz 2 + %z 2 )- (10.6)
It contains six moduli of elasticity.
A similar result is obtained for those other classes of the tetragonal system
where the natural choice of the co-ordinate axes is determined by symmetry
(I>2d> £>4, D^ h ). In the classes C 4 , S 4 , C ih , on the other hand, only the choice
of the s-axis is unique (along the axis C 4 or 5 4 ). The requirements of symmetry
then allow a further component X XX xy = -X yy yx in addition to those which
appear in (10.6). These components may be made to vanish by suitably
choosing the directions of the x and y axes, and F then reduces to the form
(10.6).
(5) Rhotnbohedral system. Let us consider the class C 3v ; we take the third-
order axis as the sr-axis, and the j-axis perpendicular to one of the vertical
planes of symmetry. In order to find the restrictions imposed on the com-
ponents of the tensor X ilc im by the presence of the axis C 3 , it is convenient
to make a formal transformation using the complex co-ordinates £ = x + iy,
r\ = x — iy, the z co-ordinate remaining unchanged. We transform the
tensor Xium to the new co-ordinate system also, so that its suffixes take the
values I, 77, z. It is easy to see that, in a rotation through 2^/3 about the
axis C3, the new co-ordinates undergo the transformation £ -> £e 2nt/3 ,
rj _> ^ e -2^f/3 ? z ->z. By symmetry only those components Xucim which are
unchanged by this transformation can be different from zero. These com-
ponents are evidently the ones whose suffixes contain £ three times, or r\
three times (since (^2^/3)3 = e 2ni = i) ? G r £ and r\ the same number of times
(since e 2^/3 e -2rf/3 = j^ i >e . X ZZZZ} X g7liv ^uw *£?**» ^£*vz> ^Hl*> ^w«-
40 Fundamental Equations §10
Furthermore, a reflection in the symmetry plane perpendicular to the jy-axis
gives the transformation x -> x, y -> -y, z -> z, or f -> rj, 77 -> £. Since
*££*z becomes A^z in this transformation, these two components must be
equal. Thus crystals of the rhombohedral system have only six moduli of
elasticity. In order to obtain an expression for the free energy, we must form
the sum \,\iicimUiicUim, in which the suffixes take the values £, t], z; since F
is to be expressed in terms of the components of the strain tensor in
the co-ordinates x, y, z, we must express in terms of these the components
in the co-ordinates £, rj, z. This is easily done by using the fact that the
components of the tensor u ik transform as the products of the corresponding
co-ordinates. For example, since
£ 2 = (x+iy) 2 = x 2 —y 2 + 2ixy,
it follows that
u ii = u %% — u yy "f" 2iu X y.
Consequently, the expression for F is found to be
F = &ZZZzU Z Z 2 + 2A iviv (u zx +U 1/V ) 2 + A^ vv {(u Xa; -Uyy) 2 + 4u X y 2 } +
+ 2X^zz(U zz + Uyy)u ZZ + A\ Zrl z{u X z 2 + Uy z 2 ) + ^^ z {(u xx - Uyy)u xz - 2u X yUy z }.
(10.7)
This contains 6 independent coefficients. A similar result is obtained for the
classes D 3 and D M , but in the classes C3 and ^6, where the choice of the x and 3/
axes remains arbitrary, requirements of symmetry allow also a non-zero value
of the difference X i ^ z -X 7j1j7jZ . This, however, can be made to vanish by a
suitable choice of the x and y axes.
(6) Hexagonal system. Let us consider the class Cq; we take the sixth-
order axis as the sr-axis, and again use the co-ordinates $ = x+iy> r\ = x — iy.
In a rotation through an angle \n about the #-axis, the co-ordinates £, r\
undergo the transformation £ -> ge^' 3 , r\ -> rje-^ 3 . Hence we see that only
those components Amm are non-zero which contain the same number of
suffixes i and r\. These are A zzzz , \ gv g v A £f „, X gr)ZZ , X gZrjZ . Other symmetry
elements in the hexagonal system give no further restrictions. There are
therefore only five moduli of elasticity. The free energy is
F = %X zzzz U ZZ 2 + 2X Ev £ v (U XX + Uyy) 2 + X iivv [(u XX -Uyy) 2 + 4u X y 2 ] +
+ 2X^ZZU ZZ (U XX + Uyy) + iX iZvZ (u xz 2 + U yz 2 ). (10.8)
It should be noticed that a deformation in the ry-plane (for which u xx ,
Uyy and u xy are non-zero) is determined by only two moduli of elasticity,
as for an isotropic body; that is, the elastic properties of a hexagonal crystal
are isotropic in the plane perpendicular to the sixth-order axis.
For this reason the choice of axis directions in this plane is unimportant and
does not affect the form of F. The expression (10.8) therefore applies to all
classes of the hexagonal system.
§10 The elastic properties of crystals 41
(7) Cubic system. We take the axes along the three fourth-order axes of
the cubic system. Since there is tetragonal symmetry (with the fourth-order
axis in the ^-direction), the number of different components of the tensor
hklm is limited to at most the following six: X xxxx , ^zzzz, ^xxzz, ^xxyy, ^xyxy,
^xzxz' Rotations through far about the x and y axes give respectively the
transformations x -> x, y -> — z, z ->y, and x -> z, y -»■ y, z -> — x. The
components listed are therefore equal in successive pairs. Thus there remain
only three different moduli of elasticity. The free energy of crystals of the
cubic system is
■** = 2^XXXX\ U XX +Uyy + Uzz ) + A XX yy[U xx Uyy-\- U XX U ZZ~\~ Uyytlzz) +
+ 2\ z y X y(ll x y 2 + U XZ 2 + Uy Z 2 ). (10.9)
We may recapitulate the number of independent parameters (elastic moduli
or angles defining the orientation of axes in the crystal) for the classes of the
various systems :
Triclinic
21
Monoclinic
13
Orthorhombic
9
Tetragonal (C4, S4, Cm)
7
Tetragonal (C 4v , D 2 d, #4, D 4h )
6
Rhombohedral (C3, 56)
7
Rhombohedral (C^v, D3, Dsa)
6
Hexagonal
5
Cubic
3
The least number of non-zero moduli that is possible by suitable choice of
the co-ordinate axes is the same for all the classes in each system :
Triclinic 18
Monoclinic 12
Orthorhombic 9
Tetragonal 6
Rhombohedral 6
Hexagonal 5
Cubic 3
All the above discussion relates, of course, to single crystals. Polycrystalline
bodies whose component crystallites are sufficiently small may be regarded
as isotropic bodies (since we are concerned with deformations in regions
large compared with the dimensions of the crystallites). Like any isotropic
body, a polycrystal has only two moduli of elasticity. It might be thought at
first sight that these moduli could be obtained from those of the individual
crystallites by simple averaging. This is not so, however. If we regard the
deformation of a polycrystal as the result of a deformation of its component
crystallites, it would in principle be necessary to solve the equations of
42 Fundamental Equations §10
equilibrium for every crystallite, taking into account the appropriate boun-
dary conditions at their surfaces of separation. Hence we see that the relation
between the elastic properties of the whole crystal and those of its component
crystallites depends on the actual form of the latter and the amount of correla-
tion of their mutual orientations. There is therefore no general relation
between the moduli of elasticity of a polycrystal and those of a single crystal
of the same substance.
The moduli of an isotropic polycrystal can be calculated with fair accuracy
from those of a single crystal only when the elastic properties of the single
crystal are nearly isotropic.f In a first approximation, the moduli of elasticity
of the polycrystal can then simply be put equal to the "isotropic part" of the
moduli of the single crystal. In the next approximation, terms appear which
are quadratic in the small "anisotropic part" of these moduli. It is found %
that these correction terms are independent of the shape of the crystallites
and of the correlation of their orientations, and can be calculated in a general
form.
Finally, let us consider the thermal expansion of crystals. In isotropic
bodies, the thermal expansion is the same in every direction, so that the
strain tensor in free thermal expansion is (see §6) um = $<x.(T— Tb)S$fc, where
a is the thermal expansion coefficient. In crystals, however, we must put
u ik = \*i k {T-T ) t (10.10)
where a^ is a tensor of rank two, symmetrical in the suffixes i and k. Let us
calculate the number of independent components of this tensor in crystals
of the various systems. The simplest way of doing this is to use the result of
tensor algebra that to every symmetrical tensor of rank two there corresponds
a tensor ellipsoid.^ It follows at once from considerations of symmetry that,
for triclinic, monoclinic and orthorhombic symmetry, the tensor ellipsoid has
three axes of different length. For tetragonal, rhombohedral and hexagonal
symmetry, on the other hand, we have an ellipsoid of revolution (with its
axis of symmetry along the axes C4, C3 and C§ respectively). Finally, for cubic
symmetry the ellipsoid becomes a sphere. An ellipsoid of three axes is
determined by three quantities, an ellipsoid of revolution by two, and a
sphere by one (the radius). Thus the number of independent components
of the tensor a^ in crystals of the various systems is as follows: triclinic,
monoclinic and orthorhombic, 3 ; tetragonal, rhombohedral and hexagonal, 2 ;
cubic, 1.
Crystals of the first three systems are said to be biaxial, and those of the
second three systems uniaxial. It should be noticed that the thermal expan-
sion of crystals of the cubic system is determined by one quantity only, i.e.
they behave in this respect as isotropic bodies.
t For a "nearly isotropic" cubic crystal (e.g.), the difference X X xxx~ X X xyy— 2X X yxv must be small.
t I. M. Lifshitz and L. N. RozentsveIg, Zhurnal eksperimental'noii teoreticheskoi fiziki 16, 967, 1946.
§ Determined by the equation «ikX{Xic — 1.
§10 The elastic properties of crystals 43
PROBLEM
Determine the Young's modulus of a cubic crystal as a function of direction.
Solution. We take the axes of co-ordinates along the three axes of the fourth order. Let
the axis of a rod cut from the crystal be in the direction of the unit vector n. The stress
tensor one in the extended rod must satisfy the following conditions: when one is multiplied
by n { , the resulting extension force must be parallel to n (condition at the ends of the rod) ;
when it is multiplied by a vector perpendicular to n, the result must be zero (condition on the
sides of the rod). Such a tensor must be of the form one = />«<«*, where p is the extension
force per unit area of the ends of the rod. Calculating the components one by means of the
expression (10.9) for the free energy! and comparing them with the formulae one = ptiitik,
we find the components of the strain tensor to be
(ci + 2c 2 )n x 2 -c 2
u xx = p- r— , u xy = pn x n y l2cz,
{ci-C2)(ci + 2c 2 )
and similarly for the remaining components. Here we have put X xxxx = c lt Xxxyy = c t ,
Axyxy — Cg.
The relative longitudinal extension of the rod is u = (dl'—dl)ldl, where dl' is given by
formula (1.2) and dxjdl — n t . For small deformations this gives u — Uijcmnjc. The Young's
modulus is determined by the coefficient of proportionality in p = Eu, and is
*- fcrSfcrfi - ^((w+^+w)]" 1 .
E has extremum values in the directions of the edges (i.e. of the co-ordinate axes) and of the
spatial diagonals of the cube.
f In calculating <r,fc, the following fact must be borne in mind. If we effect the calculation, not
directly from the formulae auc = KkimUm, but by differentiation of the expression for the free energy
with respect to the components of the tensor unc, the derivatives with respect to mk with i ^ k give
twice the values of the corresponding components o.-fc. This is because the expressions aa = dFjduuc
are meaningful only as indicating that dF = one dw.fc; in the sum <r<k du<jfc, however, the term in the
differential dune for each component with i ^ k of the symmetrical tensor «<& appears twice.
CHAPTER II
THE EQUILIBRIUM OF RODS AND PLATES
§11. The energy of a bent plate
In this chapter we shall study some particular cases of the equilibrium of
deformed bodies, and we begin with that of thin deformed plates. When we
speak of a thin plate, we mean that its thickness is small compared with its
dimensions in the other two directions. The deformations themselves are
supposed small, as before. In the present case the deformation is small if the
displacements of points in the plate are small compared with its thickness.
The general equations of equilibrium are considerably simplified when
applied to thin plates. It is more convenient, however, not to derive these
simplified equations directly from the general ones, but to calculate afresh
the free energy of a bent plate and then vary that energy.
When a plate is bent, it is stretched at some points and compressed at
others: on the convex side there is evidently an extension, which decreases
as we penetrate into the plate, finally becoming zero, after which a gradually
increasing compression is found. The plate therefore contains a neutral
surface, on which there is no extension or compression, and on opposite sides
of which the deformation has opposite signs. The neutral surface clearly
lies midway through the plate.
Fig. 2
We take a co-ordinate system with the origin on the neutral surface and the
sr-axis normal to the surface. The xy-plane is that of the undeformed plate.
We denote by £ the vertical displacement of a point on the neutral surface,
i.e. its z co-ordinate (Fig. 2). The components of its displacement in the
xy-plane are evidently of the second order of smallness relative to £, and can
therefore be put equal to zero. Thus the displacement vector for points on the
neutral surface is
u x ®> = uy^ = 0, « z <°> = i{x,y). (11.1)
44
§1 1 The energy of a bent plate 45
For further calculations it is necessary to note the following property of
the stresses in a deformed plate. Since the plate is thin, comparatively small
forces on its surface are needed to bend it. These forces are always consider-
ably less than the internal stresses caused in the deformed plate by the ex-
tension and compression of its parts. We can therefore neglect the forces P<
in the boundary condition (2.8), leaving a^n^ = 0. Since the plate is only
slightly bent, we can suppose that the normal vector n is along the #-axis.
Thus we must have on both surfaces of the plate a xz = a yz = a zz = 0. Since
the plate is thin, however, these quantities must be small within the plate
if they are zero on each surface. We therefore conclude that the components
a xz , a yz , a zz are small compared with the remaining components of the stress
tensor everywhere in the plate. We can therefore equate them to zero and
use this condition to determine the components of the strain tensor.
By the general formulae (5.13), we have
E E
o Z x = ~ u zxy a zy = - u zyi
l + cr l + o
E
° zz = Ti — ^ — ^rrft 1 ~ CT )"zz+ a ( u *x+ u vv)}-
{l + a)[l — Za)
(11.2)
Equating these expressions to zero, we obtain 8u x /8z — — 8u z /8x,
duyfdz = —8u z /dy, u zz = — o(u X x + u y y)j(\ — o). In the first two of these
equations u z can, with sufficient accuracy, be replaced by £(x, y):du x /dz =
— dt,/ dx, duyjdz = — dt,jdy> whence
u x = —zd^/dx, u y = — zdt,ldy. (11.3)
The constants of integration are put equal to zero in order to make
u x = u y = for z = 0.
Knowing u x and u y , we can determine all the components of the strain
tensor :
u xx = —zd^/dx 2 , Uyy — —zd 2 Hdy 2 , u xy = — zd^jdxdy^
a (d% &l\ (11.4)
u xz = uy z = 0, lta ._^_ + _j.
We can now calculate the free energy F per unit volume of the plate, using
the general formula (5.10). A simple calculation gives the expression
E
F= z* —
1 +
4 w^ + £?£) 2 + r^r-^i). (11 . 5)
crl2(l-<7)\a*2 dy*J [\dxdyl dx*dy*\l V '
The total free energy of the plate is obtained by integrating over the volume.
The integration over z is from — \h to + \h t where h is the thickness of the
46 The Equilibrium of Rods and Plates §12
plate, and that over x, y is over the surface of the plate. The result is that
the total free energy F pl = J*F dV of a deformed plate is
Pl 24(l-o2)j)l\dx2 + dy2) +
+2(, -' ) (® , -55)] d '*" (n - 6)
the element of area can with sufficient accuracy be written as dx dy simply,
since the deformation is small.
Having obtained the expression for the free energy, we can regard the plate
as being of infinitesimal thickness, i.e. as being a geometrical surface, since
we are interested only in the form which it takes under the action of the
applied forces, and not in the distribution of deformations inside it. The
quantity £ is then the displacement of points on the plate, regarded as a surface,
when it is bent.
§12. The equation of equilibrium for a plate
The equation of equilibrium for a plate can be derived from the condition
that its free energy is a minimum. To do so, we must calculate the variation
of the expression (11.6).
We divide the integral in (11.6) into two, and vary the two parts separately.
The first integral can be written in the form J(A £) 2 d/, where d/ = da? dy
is a surface element and A = d 2 Jdx 2 + d 2 jdy 2 is here (and in §§13, 14) the
two-dimensional Laplacian. Varying this integral, we have
8hj(A0*df= JAZA^df
= J*A£ div graded/
= Jdiv ( A £ grad S£) d/- Jgrad S£ • grad A £ d/.
All the vector operators, of course, relate to the two-dimensional co-ordinate
system (x, y). The first integral on the right can be transformed into an
integral along a closed contour enclosing the plate :f
Jdiv( AC gradSQ d/ = j> A£(n . gradS£) 61
r 2S£
= <fA£-^d/,
J on
where djdn denotes differentiation along the outward normal to the contour.
t The transformation formula for two-dimensional integrals is exactly analogous to the one for three
dimensions. The volume element dV is replaced by the surface element d/ (a scalar), and the surface
element df is replaced by a contour element dl multiplied by the vector n along the outward normal to
the contour. The integral over df is converted into one over dl by replacing the operator dfd/dxt by
fit dl. For instance, if ^ is a scalar, we have J grad <t> df = $ <£n dl.
§12 The equation of equilibrium for a plate
In the second integral we use the same transformation to obtain
Jgrad8£ . grad A£ d/ = jdiv(S£ grad AC) d/- JS£A 2 £ d/
47
= fa(n • grad A£) d/- J8£A 2 £ d/
?A^
dn
JB^dl-faAHdf
Substituting these results, we find that
^J(A0 2 d/= JHAHdf-j 8^dl+j>A^-dl. (12.1)
The transformation of the variation of the second integral in (11.6) is
somewhat more lengthy. This transformation is conveniently effected in
^U".
Fig. 3
components, and not in vector form. We have
J Wdxdy/ dx 2 dy 2 ) J
It
= 2
d% d 2 bt, d% d 2 S£ d 2 8£ dH
dx 2 dy 2
dxdy dxdy dx 2 dy 2
The integrand can be written
d/.
a /dS£ dH
B8CdH\
dx dy 2 /
d /3S£ d 2 C
a§£ dH\
dx\ dy dxdy dx dy 2 } ' dy\ dx dxdy dy dx 2 )'
i.e. as the (two-dimensional) divergence of a certain vector. The variation
can therefore be written as a contour integral:
J Wdxdy / dx 2 dy 2 ) J \dx dxdy dy dx 2 )
J \ dy dxdy dx dy 2 )
+
(12.2)
dy dxdy
where 6 is the angle between the #-axis and the normal to the contour (Fig. 3).
48 The Equilibrium of Rods and Plates §12
The derivatives of S£ with respect to x and y are expressed in terms of
its derivatives along the normal n and the tangent 1 to the contour :
B
a
a
— =
cos —
— sin —
Bx
Bn
dl
a . a a
— = sin0 l-cos0 — .
By Bn dl
Then formula (12.2) becomes
J WdxdyJ dx 2 By 2 )
r B8C( B% B 2 l B%\
= Ad/— 2 sin cos — - -sin 2 0— - -cos 2 0— - +
J Bn { BxBy Bx 2 By 2 )
r B8U I B% BH\ BH )
+ <hdl— - sin0cos0 — + (co S 20 - sin20)— - .
J Bl { \ By 2 Bx 2 / BxBy)
The second integral may be integrated by parts. Since it is taken along a
closed contour, the limits of integration are the same point, and we have
simply
r 9 ( / b 2 i a 2 £\ a 2 £ \
-^ 8i ;-|sin*cos*(- - _) + (c o*-.uW>— }.
Collecting all the above expressions and multiplying by the coefficients
shown in formula (11.6), we obtain the following final expression for the
variation of the free energy:
Eh 3 I r
r [BAl Bl [B 2 l B 2 t\
B 2 t\
+ (cos 2 0-sin 2 0)
BxBy) .
+
|_Jldz(A£ + (l-<7)(2sin0cos0
J Bn { \ BxBy
BxBy
B 2 £\\\
— sin 2 cos 2
" -«))). (12.3)
Bx 2 By 2 )
In order to derive from this the equation of equilibrium for the plate, we
must equate to zero the sum of the variation 8F and the variation 8U of the
potential energy of the plate due to the external forces acting on it. This
latter variation is minus the work done by the external forces in deforming the
§12 The equation of equilibrium for a plate 49
plate. Let P be the external force acting on the plate, per unit areaf and
normal to the surface. Then the work done by the external forces when the
points on the plate are displaced a distance S£ is JPS£ d/. Thus the condition
for the total free energy of the plate to be a minimum is
SF pl -J>S£d/=0. (12.4)
On the left-hand side of this equation we have both surface and contour
integrals. The surface integral is
The variation 8£ in this integral is arbitrary. The integral can therefore
vanish only if the coefficient of S£ is zero, i.e.
Eh*
AH-P = 0. (12.5)
12(1 -a 2 )
This is the equation of equilibrium for a plate bent by external forces acting
on it J.
The boundary conditions for this equation are obtained by equating to
zero the contour integrals in (12.3). Here various particular cases have to be
considered. Let us suppose that part of the edge of the plate is free, i.e. no
external forces act on it. Then the variations S£ and hdljdn on this part of
the edge are arbitrary, and their coefficients in the contour integrals must be
zero. This gives the equations
H(1-ct) — COS0SU10 +
dn dl{ \ dx 2 dy 2 /
+ (sin 2 0-cos 2 0) — - = 0, (12.6)
dxdy)
( d 2 l d 2 l d 2 t\
A£+(l-cr) 2 sin cos 0-^- -sin 2 0— - -cos 2 0— - = 0, (12.7)
I dxdy dx 2 dy 2 )
which must hold at all free points on the edge of the plate.
The boundary conditions (12.6) and (12.7) are very complex. Considerable
simplifications occur when the edge of the plate is clamped or supported. If
it is clamped (Fig. 4a), no vertical displacement is possible, and moreover no
f The force P may be the result of body forces (e.g. the force of gravity), and is then equal to the
integral of the body force over the thickness of the plate.
X The coefficient D = Eh 3 jl2(l — a 9 ) in this equation is called the fiexural rigidity or cylindrical
rigidity of the plate.
50 The Equilibrium of Rods and Plates §12
bending is possible at the edge. The angle through which a given part of the
edge turns from its initial position is (for small displacements £) the derivative
dt,Jdn. Thus the variations S£ and Sd£/dn must be zero at clamped edges, so
that the contour integrals in (12.3) are zero identically. The boundary con-
ditions have in this case the simple form
I = 0, dijdn = 0. (12.8)
The first of these expresses the fact that the edge of the plate undergoes no
vertical displacement in the deformation, and the second that it remains
horizontal.
m
(b)
Fig. 4
It is easy to determine the reaction forces on a plate at a point where it
is clamped. These are equal and opposite to the forces exerted by the plate
on its support. As we know from mechanics, the force in any direction is
equal to the space derivative, in that direction, of the energy. In particular,
the force exerted by the plate on its support is given by minus the derivative
of the energy with respect to the displacement £ of the edge of the plate, and
the reaction force by this derivative itself. The derivative in question, how-
ever, is just the coefficient of S£ in the second integral in (12.3). Thus the
reaction force per unit length is equal to the expression on the left of equation
(12.6) (which, of course, is not now zero), multiplied by Eh z jl2(l — a 2 ).
Similarly, the moment of the reaction forces is given by the expression on
the left of equation (12.7), multiplied by the same factor. This follows at
once from the result of mechanics that the moment of the force is equal to
the derivative of the energy with respect to the angle through which the
body turns. This angle is dt,/dn, so that the corresponding moment is given
by the coefficient of BSl/dn in the third integral in (12.3). Both these expres-
sions (that for the force and that for the moment) can be very much simplified
by virtue of the conditions (12.8). Since £ and dt,jdn are zero everywhere on
the edge of the plate, their tangential derivatives of all orders are zero also.
Using this and converting the derivatives with respect to x andjy in (12.6)
and (12.7) into those in the directions of n and 1, we obtain the following
simple expressions for the reaction force F and the reaction moment M:
Eh^ [dK dd a 2 n
F= — - + , (12.9)
12(l- CT 2)La«3 d/ drfil
Eh* 3 2 £
M = 1. (12.10)
12(l-cr 2 )d«2 v
§12 The equation of equilibrium for a plate 51
Another important case is that where the plate is supported (Fig. 4b),
i.e. the edge rests on a fixed support, but is not clamped to it. In this case
there is again no vertical displacement at the edge of the plate (i.e. on the
line where it rests on the support), but its direction can vary. Accordingly,
we have in (12.3) S£ = in the contour integral, but dS£/d» # 0. Hence
only the condition (12.7) remains valid, and not (12.6). The expression on the
left of (12.6) gives as before the reaction force at the points where the plate is
supported ; the moment of this force is zero in equilibrium. The boundary
condition (12.7) can be simplified by converting to the derivatives in the direc-
tion of n and 1 and using the fact that, since £ = everywhere on the edge, the
derivatives dljdl and d 2 £/d/ 2 are also zero. We then have the boundary
conditions in the form
dK dd dl
£ = , — + a - = 0. (12.11)
dn 2 dl dn
PROBLEMS
Problem 1 . Determine the deflection of a circular plate (of radius R) with clamped edges,
placed horizontally in a gravitational field.
Solution. We take polar co-ordinates, with the origin at the centre of the plate. The force
on unit area of the surface of the plate is P = phg. Equation (12.5) becomes A a £ = 64j3,
where j8 = 3pg(l—o*)J16h t E; positive values of £ correspond to displacements downward.
Since £ is a function of r only, we can put A = r~ l d(rd/dr)/dr. The general integral is
£ = j3r*+ar 3 +b+cr*log(r/R)+dlog(rlR). In the case in question we must put d — 0,
since log(r[R) becomes infinite at r — 0, and c = 0, since this term gives a singularity in
A £ at r — (corresponding to a force applied at the centre of the plate; see Problem 3). The
constants a and b are determined from the boundary conditions £ = 0, d£/dr = for r — R.
The result is £ = j3(.R»-r a )*.
Problem 2. The same as Problem 1, but for a plate with supported edges.
Solution. The boundary conditions (12.11) for a circular plate are
d% o-d£
£ = 0, — -+ — - = 0.
dr 2 r dr
The solution is similar to that of Problem 1, and the result is
'5 + a
/5 + cr \
£ = j8(#2_ r 2)| R2-r*\.
Problem 3. Determine the deflection of a circular plate with clamped edges when a force
/ is applied to its centre.
Solution. We have A a £ = everywhere except at the origin. Integration gives
£ = ar2 + b + cr*log(rlR),
the log r term again being omitted. The total force on the plate is equal to the force / at its
52 The Equilibrium of Rods and Plates §12
centre. The integral of A 2 £ over the surface of the plate must therefore be
12(1 -cr2)
„ f 12(1-0^)
2tt r A2£ dr = — i if.
J Eh? J
o
Hence c = 3(1— o^f/lirEh 3 . The constants a and b are determined from the boundary
conditions. The result is
3/(1 - ct2)
Problem 4. The same as Problem 3, but for a plate with supported edges.
Solution.
£ =
3/(l-<T2 )f-3 + o-_ o _ o , #
4ttM3
r3 + o- in
(R*-r*)-2r*log — \.
Ll + o- r J
Problem 5. Determine the deflection of a circular plate suspended by its centre and in a
gravitational field.
Solution. The equation for £ and its general solution are the same as in Problem 1.
Since the displacement at the centre is { = 0, we have c = 0. The constants a and b are
determined from the boundary conditions (12.6) and (12.7), which are, for circular symmetry,
dA£ d/d2£ ld£\ n d2£ ad£
= — — + -— = 0, —-+-— = 0.
r dr/ ' dr 2 r dr
dr dr\dr 2
The result is
R 3 + ct
r R 3 + a!
I = 0r2 r2 + 8#2 1og— +2ZP .
L r 1 + o-J
Problem 6. A thin layer (of thickness h) is torn off a body by external forces acting against
surface tension forces at the surface of separation. With given external forces, equilibrium is
established for a definite area of the surface separated and a definite shape of the layer
removed (Fig. 5). Derive a formula relating the surface tension to the shape of the
layer removed. f
Fig. 5
Solution. The layer removed can be regarded as a plate with one edge (the line of separa-
tion) clamped. The bending moment on the layer is given by formula (12.10). The work
done by this moment when the length of the separated surface increases by 8* is
MdSC/dx = MSxdH/dx 2
(the work of the bending force F itself is a second-order quantity). The equilibrium condition
is that this work should be equal to the change in the surface energy, i.e. to 2a.hx, where a is
| This problem was discussed by I. V. Obreimov (1930) in connection with a method which he
developed for measuring the surface tension of mica. The measurements which he made by this
method were the first direct measurements of the surface tension of solids.
§13 Longitudinal deformations of plates 53
the surface-tension coefficient, the factor 2 allowing for the creation of two free surfaces by
the separation. Thus
Eh* / 3 2 £\ 2
o 8 )!"
24(1 - a 2 ) \dxV
§13. Longitudinal deformations of plates
Longitudinal deformations occurring in the plane of the plate, and not
resulting in any bending, form a special case of deformations of thin plates.
Let us derive the equations of equilibrium for such deformations.
If the plate is sufficiently thin, the deformation may be regarded as uniform
over its thickness. The strain tensor is then a function of x and y only (the
;cy-plane being that of the plate) and is independent of z. Longitudinal
deformations of a plate are usually caused either by forces applied to its edges
or by body forces in its plane. The boundary conditions on both surfaces of
the plate are then atknjc = 0, or, since the normal vector is parallel to the
ar-axis, at z = 0, i.e. a xz = a yz = a zz = 0. It should be noticed, however,
that in the approximate theory given below these conditions continue to
hold even when the external tension forces are applied to the surfaces of the
plate, since these forces are still small compared with the resulting longi-
tudinal internal stresses (<j xx , &yy, <*xy) in the plate. Since they are zero at
both surfaces, the quantities a xz , a yz , a zz must be small throughout the
thickness of the plate, and we can therefore take them as approximately zero
everywhere in the plate.
Equating to zero the expressions (11.2), we obtain the relations
U ZZ = - 0(U ZX + Uyy)l{\- a), U XZ = Uy Z = 0. (13-1)
Substituting in the general formulae (5.13), we obtain for the non-zero com-
ponents of the stress tensor
E "
a xx — ~ 7\ u xx + aU yy)t
1 — CT Z
E
a VV = Z -AUyy+OUxx), > (13.2)
1— O*
E
°xy = ~ u xy-
1 + or
It should be noticed that the formal transformation
E -> E/(l - a 2 ), a -> a/(l - a) (13.3)
converts these expressions into those which give the relation between the
stresses a xx , a xy , a yy and the strains u xx , u yy , u zz for a plane deformation
(formulae (5.13) with u zz = 0).
Having thus eliminated the displacement u Zy we can regard the plate as a
two-dimensional medium (an "elastic plane"), of zero thickness, and take
54 The Equilibrium of Rods and Plates §13
the displacement vector u to be a two-dimensional vector with components
u x and Up. If P x and P y are the components of the external body force per
unit area of the plate, the general equations of equilibrium are
\ ox ay J
(da yx do yy \
\ ox oy I
Substituting the expressions (13.2), we obtain the equations of equilibrium in
the form
J 1 d 2 u x 1 8 2 u x 1 d 2 Uy\
Ehl h H ) + P x = 0,
ll-cx 2 dx 2 2(1 + a) dy 2 2(1 -a) dxdyj
I 1 d 2 u v 1 d 2 u v 1 d 2 u x \
Ehl - + — + -\ + Py = 0.
ll-a 2 By 2 2(1 + a) dx 2 2(1 -a) dxdyj
(13.4)
These equations can be written in the two-dimensional vector form
grad divu-^1 - <*) curl curlu = - (1 - o 2 )F/Eh, (13.5)
where all the vector operators are two-dimensional.
In particular, the equation of equilibrium in the absence of body forces is
grad divu— 1(1 — a) curl curlu = 0. (13.6)
It differs from the equation of equilibrium for a plane deformation of a body
infinite in the ^-direction (§7) only by the sign of the coefficient (in accordance
with (13.3)).f As for a plane deformation, we can introduce the stress function
defined by
<? xx = d 2 X /dy 2 , o xy = - d 2 x/dxdy, a yy = d 2 X /dx 2 , (13.7)
whereby we automatically satisfy the equations of equilibrium in the form
da xx du xy da yx da yy
1 = (J, 1 = u.
dx dy dx dy
The stress function, as before, satisfies the biharmonic equation, since for
Ax we have
A% = oxx+o-yy = E(u xx + Uyy)l(l - a) = {£/(l-cr)}divu;
this differs only by a factor from the result for a plane deformation.
It may be pointed out that the stress distribution in a plate deformed by
given forces applied to its edges is independent of the elastic constants of the
f A deformation homogeneous in the ^-direction for which a zx = a zv = a zz = everywhere is
sometimes called a state of plane stress, as distinct from a plane deformation, for which u zx = u zy =
u tz = everywhere.
§13 Longitudinal deformations of plates 55
material. For these constants appear neither in the biharmonic equation
satisfied by the stress function, nor in the formulae (13.7) which determine
the components o% from that function (nor, therefore, in the boundary
conditions at the edges of the plate).
PROBLEMS
Problem 1 . Determine the deformation of a plane disc rotating uniformly about an axis
through its centre perpendicular to its plane.
Solution. The required solution differs only in the constant coefficients from the solution
obtained in §7, Problem 5, for the plane deformation of a rotating cylinder. The radial
displacement u r = u(r) is given by the formula
u = r
8E
/3 + o- \
(tt/ 2 -4
This is the expression which gives that of §7, Problem 5, if the substitution (13.3) is made.
Problem 2. Determine the deformation of a semi-infinite plate (with a straight edge)
under the action of a concentrated force in its plane, applied to a point on the edge.
-. .^- \
Fig. 6
Solution. We take polar co-ordinates, with the angle <f> measured from the direction of
the applied force; it takes values from — (Jw+a) to frr— a, where a is the angle between the
direction of the force and the normal to the edge of the plate (Fig. 6). At every point of the
edge except that where the force is applied (the origin) we must have a^ = <r r a = 0. Using
the expressions for a^ and a r ^ obtained in §7, Problem 11, we find that the stress function
must therefore satisfy the conditions
dx 1 dx
—— = constant, --— - = constant, for <f> = -(£?! + a), (£77- -a).
or r d(f)
Both conditions are satisfied if x = rf(<f>). With this substitution, the biharmonic equation
fl d / d\ a 2 \2
[rlh-X ~dr) + ~d^) X =
gives solutions for f(<f>) of the forms sin 4>, cos <f>, <f> sin <f>, <f> cos <j>. The first two of these lead to
stresses which are zero identically. The solution which gives the correct value for the force
applied at the origin is
X = -(Flir)rtf, sincf>, a rr = -{IFIttt) cos«£, ct^ = a H = 0, (1)
where F is the force per unit thickness of the plate. For, projecting the internal stresses on
directions parallel and perpendicular to the force F, and integrating over a small semicircle
56
The Equilibrium of Rods and Plates
§13
centred at the origin (whose radius then tends to zero), we obtain
J u rr r COS (j> d<f> = —F,
a rr r sin <j> d<f> = 0,
i.e. the values required to balance the external force applied at the origin.
Formulae (1) determine the required stress distribution. It is purely radial: only a radial
compression force acts on any area perpendicular to the radius. The lines of equal stress are
the circles r = d cos <f>, which pass through the origin and whose centres lie on the line of
action of the force F (Fig. 6).
The components of the strain tensor are u„ = o„\E, u^ = — aa rr /E, u r ^ = 0. From these
we find by integration (using the expressions (1.8) for the components mic in polar co-
ordinates) the displacement vector :
IF (l-o)F
u r = \og(r/a) cos j> (f> sin <f>,
7tE 7tE
2oF 2F t , x . # (l-a)F
u A = — —sin <f) H log(r/«) sin <j> -\ (sin ^ — <f> cos <j>).
rE
TT.
E
tE
Here the constants of integration have been chosen so as to give zero displacement (trans-
lation and rotation) of the plate as a whole : an arbitrarily chosen point at a distance a from the
origin on the line of action of the force is assumed to remain fixed.
Using the solution obtained above, we can obtain the solution for any distribution of forces
acting on the edge of the plate (cf. §8). It is, of course, inapplicable in the immediate neigh-
bourhood of the origin.
Fig. 7
Problem 3. Determine the deformation of an infinite wedge-shaped plate (of angle 2a)
due to a force applied at its apex.
Solution. The stress distribution is given by formulae which differ from those of Problem
2 only in their normalisation. If the force acts along the mid-line of the wedge (Fi in Fig. 7),
we have a„ = —(Fi cos <£)/K«+i s i n 2a), a r ^ = a^ = 0. If, on the other hand, the force
acts perpendicular to this direction (F t in Fig. 7), then
a rr = — (Fz cos 4>)jr{a — \ sin 2a).
In each case the angle <f> is measured from the direction of the force.
Problem 4. Determine the deformation of a circular disc (of radius R) compressed by two
equal and opposite forces Fh applied at the ends of a diameter (Fig. 8).
Solution. The solution is obtained by superposing three internal stress distributions.
§13
Longitudinal deformations of plates
Two of these are
oVr* =
= -(IFIirrdcosfa, cr a V^i = o {1) Mi = °>
o® W . = -(2F/77T2) COS 02, aP>r* = * (2 W, = 0,
where n, & and r 2 , <f> 2 are the polar co-ordinates of an arbitrary point P with origins at A
and B respectively. These are the stresses due to a normal force F applied to a point on the
edge of a half-plane; see Problem 2. The third distribution, o(*Uk = (F/wi?)8<*, is a uniform
extension of definite intensity. For, if the point P is on the edge of the disc, we have
r x — 2R cos 4>i, *t = 2R cos ^, so that c^V/, = <* w r t r t = —F/irR. Since the directions of
r x and r 2 at this point are perpendicular, we see that the first two stress distributions give
a uniform compression on the edge of the disc. These forces can be just balanced by the
uniform tension given by the third system, so that the edge of the disc is free from stress, as it
should be.
Fig. 8
Problem 5. Determine the stress distribution in an infinite sheet with a circular aperture
(of radius R) under uniform tension.
Solution. The uniform tension of a continuous sheet corresponds to stresses o^xx = T,
ff(0) v» = <* w xy = 0, where T is the tension force. These in turn correspond to the stress
function x (0) — \Ty* = %Tr* sin 2 ^ = J7V*(1 —cos 2<f>). When there is a circular aperture
(with the centre as the origin of polar co-ordinates r, <f>), we seek the stress function in the
form x = X (0) +X (I) , X (1) =f(.i r )+F(r) cos 2<f>. The integral of the biharmonic equation which
is independent of <f> is of the form /(r) = ar* log r+br'+c log r, and in the integral pro-
portional to cos 2<f> we have F(r) = d^+ei^+glr 2 . The constants are determined by the
conditions o^Uk = for r = oo and a rr = c f ^ = for r = R. The result is
x (i) = iT R^-logr+ (l - |L) cos2^J,
and the stress distribution is given by
3i?2x
— J cos 20},
o*-»r{i+--(i+— Jco.2*J,
0^.-17(1+—-—) sin 2f
In particular, at the edge of the aperture we have <* K
o^q = 3T, i.e. three times the stress at infinity (cf. §?, Problem 12)
"ft"
T(l —2 cos 24), and for <f> = ±frr,
58 The Equilibrium of Rods and Plates §14
§14. Large deflections of plates
The theory of the bending of thin plates given in §§11-13 is applicable only
to fairly small deflections. Anticipating the result given below, it may be
mentioned here that the condition for that theory to be applicable is that the
deflection £ is small compared with the thickness h of the plate. Let us now
derive the equations of equilibrium for a plate undergoing large deflections.
The deflection £ is not now supposed small compared with h. It should be
emphasised, however, that the deformation itself must still be small, in the
sense that the components of the strain tensor must be small. In practice,
this usually implies the condition £ <^ /, i.e. the deflection must be small
compared with the dimension / of the plate.
The bending of a plate in general involves a stretching of it.f For small
deflections this stretching can be neglected. For large deflections, however,
this is not possible ; there is therefore no neutral surface in a plate undergoing
large deflections. The existence of a stretching which accompanies the
bending is peculiar to plates, and distinguishes them from thin rods, which
can undergo large deflections without any general stretching. This property
of plates is a purely geometrical one. For example, let a flat circular plate be
bent into a segment of a spherical surface. If the bending is such that the
circumference of the plate remains constant, its diameter must increase. If the
diameter is constant, on the other hand, the circumference must be reduced.
The energy (11.6), which may be called the pure bending energy, is only
the part of the total energy which arises from the non-uniformity of the
tension and compression through the thickness of the plate, in the absence
of any general stretching. The total energy includes also a part due to this
general stretching; this may be called the stretching energy.
Deformations consisting of pure bending and pure stretching have been
considered in §§11-13. We can therefore use the results obtained in these
sections. It is not necessary to consider the structure of the plate across its
thickness, and we can regard it as a two-dimensional surface of negligible
thickness.
We first derive an expression for the strain tensor pertaining to the stretch-
ing of a plate (regarded as a surface) which is simultaneously bent and
stretched in its plane. Let u be the two-dimensional displacement vector
(with components u x , u y ) for pure stretching; £, as before, denotes the trans-
verse displacement in bending. Then the element of length d/ = y^dx 2 + dy 2 )
of the undeformed plate is transformed by the deformation into an element
dl', whose square is given by d/' 2 = (dx + du x ) 2 + (dy + du y ) 2 + dl 2 . Putting
here du x = (du x /dx) dx + (du x /dy) dy, and similarly for du y and d£, we
obtain to within higher-order terms d/' 2 = dl 2 + 2u afi dx 0l dx fi , where the
two-dimensional strain tensor is defined as
i/du a du B \ i dt, dt
— ( — -+ — -1+ • (14.1)
2\dx s dxj 2dx a dx#
u,
*B
f An exception is, for instance, the bending of a flat plate into a cylindrical surface.
§14 Large deflections of plates 59
(In this and the following sections, Greek suffixes take the two values x and_y ;
as usual, summation over repeated suffixes is understood.) The terms quad-
ratic in the derivatives of u a are here omitted; the same cannot, of course, be
done with the derivatives of £, since there are no corresponding first-order
terms.
The stress tensor <r a/? due to the stretching of the plate is given by formula
(13.2), in which u afi must be replaced by the total strain tensor given by
formula (14.1). The pure bending energy is given by formula (11.6), and can
be written J Yi(£) dx dy, where *Fi(£) denotes the integrand in (11.6). The
stretching energy per unit volume of the plate is, by the general formulae,
\u afi a ap . The energy per unit surface area is obtained by multiplying by h,
so that the total stretching energy can be written J T 2 (« a/f ) d/, where
T 2 = \hu^a^. (14.2)
Thus the total free energy of a plate undergoing large deflections is
F P i= j{Ti(0 + T 2 ( Wa/? )}d/. (14.3)
Before deriving the equations of equilibrium, let us estimate the relative
magnitude of the two parts of the energy. The first derivatives of £ are of
the order of £//, where / is the dimension of the plate, and the second deriva-
tives are of the order of £/Z 2 . Hence we see from (11.6) that Ti ~ M 3 £ 2 // 4 .
The order of magnitude of the tensor components u afi is £ 2 // 2 , and so
T 2 ~ £A£ 4 // 4 . A comparison shows that the neglect of T 2 in the approximate
theory of the bending of plates is valid only if £ 2 <^ A 2 .
The condition of minimum energy is 8F+ SU = 0, where U is the poten-
tial energy in the field of the external forces. We shall suppose that the
external stretching forces, if any, can be neglected in comparison with the
bending forces. (This is always valid unless the stretching forces are very
large, since a thin plate is much more easily bent than stretched.) Then we
have for 8U the same expression as in §12: 8U = -J"PS£ d/, where P is the
external force per unit area of the plate. The variation of the integral J Ti d/
has already been calculated in §12, and is
r ew r
The contour integrals in (12.3) are omitted, since they give only the boundary
conditions on the equation of equilibrium, and not that equation itself, which
is of interest here.
Finally, let us calculate the variation of the integral J Y 2 d/. The variation
must be taken both with respect to the components of the vector u and with
respect to £. We have
r raT 2
SPF 2 d/= — Su^df.
J J ou a
The derivatives of the free energy per unit volume with respect to u xp are
60 The Equilibrium of Rods and Plates §14
a a/} ; hence d^l^Uafi = ha afi . Substituting also for u a/} the expression (14.1),
we obtain
8JV 2 df=hj* a/i 8u a0 df
r (B8u a dhu B dt, 3S£ 3S£ d£\
= pUJ — -+ — -+——+— —4/;
J l dxo dxg, dx a dxo dx a dxj
or, by the symmetry of a a/? ,
(d8u a dSC 3£
Cg 0X0
Integrating by parts, we obtain
da^ . 3/ 3£
*0
8 [V 2 d/ = h LJ^l + ^L ■£-} d/.
J J I oa^ 3*^ 3#J
ts, we obtain
8 |V 2 d/ = -h f |^8« a + J-( ff JL\ 8 (;\ d /.
The contour integrals along the circumference of the plate are again omitted.
Collecting the above results, we have
8F vl + 8U = (U AH-h — (o aS -^\-p)8t,-h-^8u} d/ = 0.
P JLll2(l-o*) dxA^dxJ I dx B a l J
In order that this relation should be satisfied identically, the coefficients of
81 and 8u x must each be zero. Thus we obtain the equations
M3 3 / 3£
AH-h— '
12(1 -a 2 ) dx
do^dx, = 0. (14.5)
The unknown functions here are the two components u Xl u y of the vector
u and the transverse displacement £. The solution of the equations gives both
the form of the bent plate (i.e. the function £(#, y)) and the extension resulting
from the bending. Equations (14.4) and (14.5) can be somewhat simplified
by introducing the function x related to a afi by (13.7). Equation (14.4) then
becomes
/3 2 v 3 2 £ 3 2 v 3 2 £ 3 2 v 3 2 £ \
-hi— £— + ——- 2— *-— I =P. (14.6)
\ 3V 2 dx 2 dx 2 dy 2 dxdy dxdyf
EW> (d 2 x d 2 £ d 2 X dH &X d 2 £
12(1 — a 2 ) \ dy 2 dx 2 dx 2 dy 2 dxdy dxdy/
Equations (14.5) are satisfied automatically by the expressions (13.7). Hence
another equation is needed ; this can be obtained by eliminating u a from the
relations (13.7) and (13.2).
To do this, we proceed as follows. We express w a/ff in terms of a a/? , obtaining
from (13.2)
U X X = (VXX— 0-Oyy)IE, Uyy — {Oyy— OO XX )/E, U X y = (1 + <j)(J X y/E.
§14 Large deflections of plates 61
Substituting here the expression (14.1) for u a/s , and (13.7) for o a p, we find
the equations
dx + 2\dx) ~ E\dy 2 ° dx 2 )'
dy + 2\dy) E\dx 2 ° dy 2 )'
du x du y d^dt, _ 2(1 + a) d 2 x
dy dx dx dy E dxdy
We take d 2 jdy 2 of the first, d 2 /dx 2 of the second, — d 2 /dxdy of the third, and
add. The terms in u x and u y then cancel, and we have the equation
2 jdH d% i d 2 £\ 2 \ _
\ dx 2 dy 2 \dxdyl )
Equations (14.6) and (14.7) form a complete system of equations for large
deflections of thin plates (A. Foppl 1907). These equations are very compli-
cated, and cannot be solved exactly, even in very simple cases. It should be
noticed that they are non-linear.
We may mention briefly a particular case of deformations of thin plates,
that of membranes. A membrane is a thin plate subject to large external
stretching forces applied at its circumference. In this case we can neglect
the additional longitudinal stresses caused by bending of the plate, and
therefore suppose that the components of the tensor cr a/? are simply equal to
the constant external stretching forces. In equation (14.4) we can then
neglect the first term in comparison with the second, and we obtain the
equation of equilibrium
hcr a ,~?^-+P=0 t (14.8)
with the boundary condition that £ = at the edge of the membrane. This
is a linear equation. The case of isotropic stretching, when the extension of
the membrane is the same in all directions, is particularly simple. Let T be
the absolute magnitude of the stretching force per unit length of the edge of
the membrane. Then ha afi = T8 afi , and we obtain the equation of equili-
brium in the form
TAt + P = 0. (14.9)
PROBLEMS
Problem 1. Determine the deflection of a plate as a function of the force on it when
Solution. An estimate of the terms in equation (14.7) shows that x ~ EC*. For { ^> h,
the first term in (14.6) is small compared with the second, which is of the order of magnitude
htxll* ~ EH* 1 1* (I being the dimension of the plate). If this is comparable with the external
62 The Equilibrium of Rods and Plates §15
force P, we have £ ~ (l l P{Eh)l. Hence, in particular, we see that £ is proportional to the
cube root of the force.
Problem 2. Determine the deformation of a circular membrane (of radius R) placed
horizontally in a gravitational field.
Solution. We have P = pgh; in polar co-ordinates, (14.9) becomes
r dr \ or!
1 d / d£\ P gh_
T'
The solution finite for r — and zero for r = R is £ = pgh(R 2 —r 2 )/4T.
§15. Deformations of shells
In discussing hitherto the deformations of thin plates, we have always
assumed that the plate is flat in its undeformed state. However, deformations
of plates which are curved in the undeformed state (called shells) have
properties which are fundamentally different from those of the deformations
of flat plates.
The stretching which accompanies the bending of a flat plate is a second-
order effect in comparison with the bending deflection itself. This is seen,
for example, from the fact that the strain tensor (14.1), which gives this
stretching, is quadratic in £. The situation is entirely different in the defor-
mation of shells : here the stretching is a first-order effect, and therefore is
important even for small bending deflections. This property is most easily
seen from a simple example, that of the uniform stretching of a spherical
shell. If every point undergoes the same radial displacement £, the length
of the equator increases by Inl,. The relative extension is IttX^IttR — t,jR f
and hence the strain tensor also is proportional to the first power of £. This
effect tends to zero as R -> oo, i.e. as the curvature tends to zero, and is
therefore due to the curvature of the shell.
Let i? be the order of magnitude of the radius of curvature of the shell,
which is usually of the same order as its dimension. Then the strain tensor
for the stretching which accompanies the bending is of the order of £jR,
the corresponding stress tensor is ~ E£jR, and the deformation energy per
unit area is, by (14.2), of the order of Eh{ljRf. The pure bending energy, on
the other hand, is of the order of Eh^jR^, as before. We see that the ratio of
the two is of the order of (R/h) 2 , i.e. it is very large. It should be emphasised
that this is true whatever the ratio of the bending deflection £ to the thickness
h, whereas in the bending of flat plates the stretching was important only
for £ £ h.
In some cases there may be a special type of bending of the shell in
which no stretching occurs. For example, a cylindrical shell (open at both
ends) can be deformed without stretching if all the generators remain parallel
(i.e. if the shell is, as it were, compressed along some generator). Such
deformations without stretching are geometrically possible if the shell has
free edges (i.e. is not closed) or if it is closed but its curvature has opposite
§15 Deformations of shells 63
signs at different points. For example, a closed spherical shell cannot be
bent without being stretched, but if a hole is cut in it (the edge of the hole
not being fixed), then such a deformation becomes possible. Since the pure
bending energy is small compared with the stretching energy, it is clear that,
if any given shell permits deformation without stretching, then such defor-
mations will, in general, actually occur when arbitrary external forces act on
the shell. The requirement that the bending is unaccompanied by stretching
places considerable restrictions on the possible displacements u u . These
restrictions are purely geometrical, and can be expressed as differential
equations, which must be contained in the complete system of equilibrium
equations for such deformations. We shall not pause to discuss this question
further.
If, however, the deformation of the shell involves stretching, then the
tensile stresses are in general large compared with the bending stresses,
which may be neglected. Shells for which this is done are called membranes.
The stretching energy of a shell can be calculated as the integral
F P i = ^hju xfi a afi df (15.1)
taken over the surface. Here w a/ff (oc, /? = 1, 2) is the two-dimensional strain
tensor in the appropriate curvilinear co-ordinates, and the stress tensor a a p
is related to u a/3 by formulae (13.2), which can be written, in two-dimensional
tensor notation, as
(V = E[(l - a)u ap + o8 a/) u Yy ]l(l - CT 2). (15.2)
A case requiring special consideration is that where the shell is subjected
to the action of forces applied to points or lines on the surface and directed
through the shell. These may be, in particular, the reaction forces on the shell
at points (or lines) where it is fixed. The concentrated forces result in a
bending of the shell in small regions near the points where they are applied;
let d be the dimension of such a region for a force /applied at a point (so that
its area is of the order of d 2 ). Since the deflection £ varies considerably over a
distance d, the bending energy per unit area is of the order of Eh z l 2 /d' i , and the
total bending energy (over an area ~ d 2 ) is of the order oiEtfitpjd 2 . The strain
tensor for the stretching is again ~ £/i?, and the total stretching energy due to
the concentrated forces is ~ Eh£, 2 d 2 /R 2 . Since the bending energy increases
and the stretching energy decreases with decreasing d, it is clear that both
energies must be taken into account in determining the deformation near the
point of application of the forces. The size d of the region of bending is given
in order of magnitude by the condition that the sum of these energies is a
minimum, whence
d ~ y/(hR). (15.3)
64
The Equilibrium of Rods and Plates
§15
The energy ~ Eh 2 £, 2 /R. Varying this with respect to £ and equating the result
to the work done by the force/, we find the deflection £ ~ fR/Eh 2 .
However, if the forces acting on the shell are sufficiently large, the shape of
the shell may be considerably changed by bulges which form in it. The
determination of the deformation as a function of the applied loads requires
special investigation in this unusual case.f
Let a convex shell (with edges fixed in such a way that it is geometrically
rigid) be subjected to the action of a large concentrated force / along the in-
ward normal. For simplicity we shall assume that the shell is part of a sphere
of radius R. The region of the bulge will be a spherical cap which is almost a
mirror image of its original shape (Fig. 9 shows a meridional section of the
shell). The problem is to determine the size of the bulge as a function of the
force.
The major part of the elastic energy is concentrated in a narrow strip near
the edge of the bulge, where the bending of the shell is relatively large ; we
shall call this the bending strip and denote its width by d. This energy may be
estimated, assuming that the radius r of the bulge region is much less than R,
so that the angle a <^ 1 (Fig. 9). Then r — R sin a ~ Ra, and the depth of the
Fig. 9
bulge H = 2R(1 — cos a) ~ jRa 2 . Let £ denote the displacement of points on
the shell in the bending strip. Just as previously, we find that the energies of
bending along the meridian and of stretching along the circle of latitude:}: per
unit surface area are respectively, in order of magnitude, Eh 3 t, 2 ld 4 and
t The results given below are due to A. V. Pogorelov (1960). A more precise analysis of the problem
together with some similar ones is given in his book Teoriya obolochek pri zakriticheskikh deformatsiyakh
{Theory of Shells at Supercritical Deformations), Moscow 1965.
% The curvature of the shell does not affect the bending along the meridian in the first approximation,
so that this bending occurs without any general stretching along the meridian, as in the cylindrical
bending of a flat plate.
§15 Deformations of shells 65
Ehl 2 /R 2 . The order of magnitude of the displacement £ is in this case deter-
mined geometrically: the direction of the meridian changes by an angle ~a
over the width d, and so £ ~ <xd ~ rd\R. Multiplying by the area of the bend-
ing strip (~n/), we obtain the energies Eh^jR 2 d and EhdW/R*. The condi-
tion for their sum to be a minimum again gives d~-\/{hR), and the total
elastic energy is then ~ Er 3 (h/R) 5/2 , orf
constant x Eh™. H^ l2 jR. (15.4)
In this derivation it has been assumed that d<^r; formula (15.4) is therefore
valid if the condition^
Rhjr*<^\ (15.5)
holds.
The required relation between the depth of the bulge H and the applied
force / is obtained by equating/ to the derivative of the energy (15.4) with
respect to H. Thus we find
■ H~f 2 R 2 \E 2 hs. ■ . . -, , : ; . (15.6)
It should be noticed that this relation is non-linear.
Finally, let the deformation (bulge) of the shell occur under a uniform
external pressure^). In this case the work done ispAV, where AV~ Hr 2 ~ H 2 R
is the change in the volume within the shell when the bulge occurs. Equating
to zero the derivative with respect to H of the total free energy (the difference
between the elastic energy (15.4) and this work), we obtain
H~h*E 2 lR*p 2 . (15.7)
The inverse variation (H increasing when p decreases) shows that in this case
the bulge is unstable. The value of H given by formula (15.7) corresponds to
unstable equilibrium for a given p: bulges with larger values of H grow of
their own accord, while smaller ones shrink (it is easy to verify that (15.7)
corresponds to a maximum and not a minimum of the total free energy).
There is a critical value p cr of the external load beyond which even small
changes in the shape of the shell increases in size spontaneously. This value
may be defined as that which gives H~h in (15.7):
p CT ~Eh 2 IR 2 . (15.8)
We shall add to the above brief account of shell theory only a few simple
examples in the following Problems.
t A more accurate calculation shows that the constant coefficient is 1.2 (1— <r 2 ) _3 ' 4 .
t When a bulge is formed, the outer layers of a spherical segment become the inner ones and are
therefore compressed, while the inner layers become the outer ones and are stretched. The relative
extension (or compression) ~ hjR, and so the corresponding total energy in the region of the bulge
~ E(h/R) 2 kr*. With the condition (15.5) it is in fact small in comparison with the energy (15.4) in
the bending strip.
-5*
66
The Equilibrium of Rods and Plates
§15
PROBLEMS
Problem 1 . Derive the equations of equilibrium for a spherical shell (of radius R) deformed
symmetrically about an axis through its centre.
Solution. We take as two-dimensional co-ordinates on the surface of the shell the angles
0, <f> in a system of spherical polar co-ordinates, whose origin is at the centre of the sphere and
polar axis along the axis of symmetry of the deformed shell.
Let P r be the external radial force per unit surface area. This force must be balanced by a
radial resultant of internal stresses acting tangentially on an element of the shell. The con-
dition is
h(a,,+ a e0 )IR = P r
(1)
This equation is exactly analogous to Laplace's equation for the pressure difference between
two media caused by surface tension at the surface of separation.
Next, let Qz(0) be the resultant of all external forces on the part of the shell lying above the
co-latitude 9; this resultant is along the polar axis. The force Qz(9) must be balanced by the
projection on the polar axis of the stresses lirRhagg sin 9 acting on the cross-section 2irRh sin 9
of the shell at that latitude. Hence
2TrRha ee sin 2 = Q z (6).
(2)
Equations (1) and (2) determine the stress distribution, and the strain tensor is then given
by the formulae
Finally, the displacement vector is obtained from the equations
= 0.
u ee
1 / du e
— I Yu r
R\dd
)■
1
U d( p — — (u d cotd + u r ).
R
(3)
(4)
Problem 2. Determine the deformation under its own weight of a hemispherical shell
convex upwards, the edge of which moves freely on a horizontal support (Fig. 10).
Fig. 10
Solution. We have P r = — pgh cos 9, Q z = —2iTR 2 pgh(l —cos 9); Q z is the total weight
of the shell above the circle of co-latitude 9. From (1) and (2) of Problem 1 we find
a eo - -
Rpg
1 + COS 0'
a rr
= Rpg\ cos 01.
^\1+COS0 /
From (3) we calculate u^ and ugg, and then obtain ug and u r from (4); the constant in the
integration of the first equation (4) is chosen so that for 9 = \n we have Ug — 0. The result
Ug =
R 2 pg(l + a){ COS0
E
1 + cos 6
+ log(l + cos0) sin0,
U r = (1
2+.
/ i
-cos 0— cos log(l + COS 0) J.
E { 1 + ff
The value of u T for = \ti gives the horizontal displacement of the support.
§15
Deformations of Shells
67
Problem 3. Determine the deformation of a hemispherical shell with clamped edges,
convex downwards and filled with liquid (Fig. 11); the weight of the shell itself can be
neglected in comparison with that of the liquid.
Fig. 11
Solution. We have
P r = P ogR cos d, P* = 0,
f 2
Q z = 2ttR? P r cos sin d0 = -nR?pog(l - cos30),
where p is the density of the liquid. We find from (1) and (2) of Problem 1
a ee
R 2 pog 1 — cos 3
~3h sin 2 '
The displacements are
R 2 Pog -l + 3cos0-2cos 3
2h
sin 2
R?pog(l + a) . r cos0
u d = : sin
U r =
3Eh
R3f >0g(l + cr)
3Eh
r cos0 I
— -+log(l + cos0) ,
LI + cos 6 J
r 3 cos 1
cos01og(l + cos0)-l + .
For 9 — $n, u r is not zero as it should be. This means that the shell is actually so severely bent
near the clamped edge that the above solution is invalid.
Problem 4. A shell in the form of a spherical cap rests on a fixed support (Fig. 12). Deter-
mine the bending resulting from the weight Q of the shell.
Fig. 12
68 The Equilibrium of Rods and Plates §16
Solution. The main deformation occurs near the edge, which is bent as shown by the
dashed line in Fig. 12. The displacement ug is small compared with the radial displacement
M r = £• Since £ decreases rapidly as we move away from the supported edge, the deformation
can be regarded as that of a long flat plate (of length 2irR sin a). This deformation is composed
of a bending and a stretching of the plate. The relative extension at each point is Z/R (R
being the radius of the shell), and therefore the stretching energy is Et?l2R* per unit volume.
Using as the independent variable the distance x from the line of support, we have for the
total stretching energy
Eh
eh r
jPipi = 2ttR sin a—— £ 2 dx,
The bending energy is
Effi f/d2A2
i<2 P i = 2ttR sin a — I dx.
24(1-0*) J \dW
Varying the sum F pl = F lp i +F 2 p i with respect to £, we obtain
d*£ 12(1-0*)
— -+— - X = 0.
d*4 h*R*
For x -*■ oo, £ must tend to zero, and for * = we must have the boundary conditions of
zero moment of the forces (£" = 0) and equality of the normal force and the corresponding
component of the force of gravity:
27ri?sina £'" = Qcosa.
12(1 - a*) *
The solution which satisfies these conditions is £ = Ae~ KX cos kx, where
_ r3(l-o2)-|i/4 _ 0cot a r3i?2(i-o-2)-|i/4
L A 2 #2 J ' A " Eh [ 8W J '
The bending of the shell is
d = £(0) cos a = A cos a.
§16. Torsion of rods
Let us now consider the deformation of thin rods. This differs from all
the cases hitherto considered, in that the displacement vector u may be large
even for small strains, i.e. when the tensor u^ is small.f For example, when
a long thin rod is slightly bent, its ends may move a considerable distance,
even though the relative displacements of neighbouring points in the rod
are small.
There are two types of deformation of a rod which may be accompanied by
a large displacement of certain parts of it. One of these consists in bending
f The only exception is a simple extension of a rod without change of shape, in which case the
vector u is always small if the tensor wjfc is small, i.e. if the extension is small.
§16 Torsion of rods 69
the rod, and the other in twisting it. We shall begin by considering the latter
case.
A torsional deformation is one in which, although the rod remains straight,
each transverse section is rotated through some angle relative to those below
it. If the rod is long, even a slight torsion causes sufficiently distant cross-
sections to turn through large angles. The generators on the sides of the rod,
which are parallel to its axis, become helical in form under torsion.
Let us consider a thin straight rod of arbitrary cross-section. We take a
co-ordinate system with the sr-axis along the axis of the rod and the origin
somewhere inside the rod. We use also the torsion angle t, which is the angle
of rotation per unit length of the rod. This means that two neighbouring
cross-sections at a distance dsr will rotate through a relative angle d^ = t dz
(so that t = d<f>(dz). The torsional deformation itself, i.e. the relative dis-
placement of adjoining parts of the rod, is assumed small. The condition
for this to be so is that the relative angle turned through by cross-sections
of the rod at a distance apart of the order of its transverse dimension R is
small, i.e.
rR < 1. (16.1)
Let us examine a small portion of the length of the rod near the origin, and
determine the displacements u of the points of the rod in that portion. As
the undisplaced cross-section we take that given by the «y-plane. When a
radius vector r turns through a small angle 8<f>, the displacement of its end
is given by
8r = 5«J>Xr, (16.2)
where S<|> is a vector whose magnitude is the angle of rotation and whose
direction is that of the axis of rotation. In the present case, the rotation is
about the #-axis, and for points of co-ordinate z the angle of rotation relative
to the ary-plane is rz (since t can be regarded as a constant in some region
near the origin). Then formula (16.2) gives for the components u x , u y of the
displacement vector
u x = -Tzy, u y = tzx. (16.3)
When the rod is twisted, the points in it in general undergo a displacement
along the sr-axis also. Since for t = this displacement is zero, it may be
supposed proportional to t when t is small. Thus
u z = nls( X) y), (16.4)
where i/r(#, y) is some function of x and y, called the torsion function. As a
result of the deformation described by formulae (16.3) and (16.4), each cross-
section of the rod rotates about the #-axis, and also becomes curved instead
of plane. It should be noted that, by taking the origin at a particular point in
the #y-plane, we "fix" a certain point in the cross-section of the rod in such a
70 The Equilibrium of Rods and Plates §16
way that it cannot move in that plane (but it can move in the ^-direction).
A different choice of origin would not, of course, affect the torsional deforma-
tion itself, but would give only an unimportant displacement of the rod as a
whole.
Knowing u, we can find the components of the strain tensor. Since u is
small in the region under consideration, we can use the formula
uoc = \{duijdxk+du]cjdxi).
The result is
Uxx — u yy — u xy — u zz — 0,
Uxz = $t(— - yj, u yz = \r\— + x\. (16.5)
It should be noticed that uu = ; in other words, torsion does not result in
a change in volume, i.e. it is a pure shear deformation.
For the components of the stress tensor we find
&xx — a yy = a zz = a xy = 0,
a xz = 2fiu zz = firi- yu o yz = 2[xu yz = /*rl— + *J. (16.6)
Here it is more convenient to use the modulus of rigidity /x in place of E and
a. Since only a xz and a yz are different from zero, the general equations
of equilibrium dancjdxjc = reduce to
i^+^fl = 0. (16.7)
dx dy
Substituting (16.6), we find that the torsion function must satisfy the equation
A«A = 0, (16.8)
where A is the two-dimensional Laplacian.
It is rather more convenient, however, to use a different auxiliary function
X ( x > y)> defined by
a xz = 2firdxldy, Oyz = -2firdx/dx; (16.9)
this function satisfies more convenient boundary conditions on the circum-
ference of the rod (see below). Comparing (16.9) and (16.6), we obtain
*± = y + 2 d I, d l=- x -2% (16.10)
dx dy dy dx
Differentiating the first of these with respect to y, the second with respect to
§16 Torsion of rods 71
x, and subtracting, we obtain for the function x the equation
AX=-1- (16.11)
To determine the boundary conditions on the surface of the rod, we note
that, since the rod is thin, the external forces on its sides must be small com-
pared with the internal stresses in the rod, and can therefore be put equal to
zero in seeking the boundary conditions. This fact is exactly analogous to
what we found in discussing the bending of thin plates. Thus we must have
oiknjc = on the sides of the rod; since the ^-direction is along the axis,
rig = 0, and this equation becomes
aZxHx + Ozytly = 0.
Substituting (16.9), we obtain
d X d X ~
— n x n y — "•
dy dx
The components of the vector normal to a plane contour (the circumference
of the rod) are n x — — dyjdl, n y = dxjdl, where x and y are co-ordinates
of points on the contour and d/ is an element of arc. Thus we have
dx ^x
— dx H dy = dx = 0,
dx dy
whence x — constant, i.e. x ls constant on the circumference. Since only
the derivatives of the function x appear in the definitions (16.9), it is clear
that any constant may be added to x- If the cross-section is singly connected,
we can therefore use, without loss of generality, the boundary condition
X = (16.12)
on equation (16.1 l).f
Fig. 13
For a multiply connected cross-section, however, x v*d\ have different
constant values on each of the closed curves bounding the cross-section.
f The problem of determining the torsion deformation from equation (16.11) with the boundary
condition (16.12) is formally identical with that of determining the bending of a uniformly loaded
plane membrane from equation (14.9).
It is useful to note also an analogy with fluid mechanics: an equation of the form (16.11) determines
the velocity distribution v(x, y) for a viscous fluid in a pipe, and the boundary condition (16.12)
corresponds to the condition v = at the fixed walls of the pipe (see Fluid Mechanics, §17).
72 The Equilibrium of Rods and Plates §16
Hence we can put x = on only one of these curves, for instance the outer-
most (Co in Fig. 13). The values of x on the remaining bounding curves are
found from conditions which are a consequence of the one-valuedness of the
displacement u z = rifj(x,y) as a function of the co-ordinates. For, since the
torsion function tfj(x, y) is one- valued, the integral of its differential d^r round
a closed contour must be zero. Using the relations (16.10), we therefore
have
j**-§Q**+%*y)
= — 2 (1) I — dy dx\ — 2 (p (# dy—y dx)
= 0,
or
— d/= -S, (16.13)
dn
where dx/dn is the derivative of the function x along the outward normal
to the curve, and S the area enclosed by the curve. Applying (16.13) to each
of the closed curves C\, C2, ••• , we obtain the required conditions.
Let us determine the free energy of a rod under torsion. The energy per
unit volume is
F = \0UcUilc = O xz U xz + OyzUy Z = ( O xz 2 + Oy z 2 )/2fl
or, substituting (16.9),
where grad denotes the two-dimensional gradient. The torsional energy
per unit length of the rod is obtained by integrating over the cross-section
of the rod, i.e. it is £CV 2 , where the constant C = 4/x J (grad x) 2 d/, and is
called the torsional rigidity of the rod. The total elastic energy of the rod is
equal to the integral
F rod = i J Ct2 d*, (16.14)
taken along its length.
Putting
(gradx) 2 = div(xgradx)-xAx = divfo grad*) + x
and transforming the integral of the first term into one along the circumference
of the rod, we obtain
C = 4/x^x^d/+4^Jxd/. (16.15)
§16 Torsion of rods 73
If the cross-section is singly connected, the first term vanishes by the
boundary condition x = 0> leaving
C = Aii J" x dx d y- (16.16)
For a multiply connected cross-section (Fig. 13), we put x = on the outer
boundary C and denote by xfr the constant values of x on the inner boun-
daries Ck, obtaining by (16.13)
c = 4/*2x*s*+4/* jx <** 4y; ( 16 - 17 )
*
it should be remembered that, in integrating in the first term in (16.15), we go
anti-clockwise round the contour Co and clockwise round all the others.
Let us consider now a more usual case of torsion, where one of the ends of
the rod is held fixed and the external forces are applied only to the other end.
These forces are such that they cause only a twisting of the rod, and no other
deformation such as bending. In other words, they form a couple which twists
the rod about its axis. The moment of this couple will be denoted by M.
We should expect that, in such a case, the torsion angle t is constant
along the rod. This can be seen, for example, from the condition that the free
energy of the rod is a minimum in equilibrium. The total energy of a de-
formed rod is equal to the sum F T0 <i+ U, where U is the potential energy
due to the action of the external forces. Substituting in (16.14) r = df/dz
and varying with respect to the angle <f>, we find
or, integrating by parts,
_ f C-^8<f>dz + 8U+[Cr8<f>] = 0.
J dsr
The last term on the left is the difference of the values at the limits of inte-
gration, i.e. at the ends of the rod. One of these ends, say the lower one, is
fixed, so that 8cf> = there. The variation 8 U of the potential energy is
minus the work done by the external forces in rotation through an angle 8(f).
As we know from mechanics, the work done by a couple in such a rotation
is equal to the product M8<f> of the angle of rotation and the moment of the
couple. Since there are no other external forces, 8U = -M8<f>, and we
have
C^8<f>dz+[8<f>(-M+Cr)] = 0. (16.18)
i<
dz
The second term on the left has its value at the upper end of the rod. In the
74 The Equilibrium of Rods and Plates §16
integral over z, the variation 8<f> is arbitrary, and so we must have
Cdr/dz = 0,
i.e.
t = constant. (16.19)
Thus the torsion angle is constant along the rod. The total angle of rotation
of the upper end of the rod relative to the lower end is rl, where / is the length
of the rod.
In equation (16.18), the second term also must be zero, and we obtain the
following expression for the constant torsion angle :
r = M/C. (16.20)
PROBLEMS
Problem 1. Determine the torsional rigidity of a rod whose cross-section is a circle of
radius R.
Solution. The solutions of Problems 1-4 are formally identical with those of problems of
the motion of a viscous fluid in a pipe of corresponding cross-section (see the last footnote
to this section). The discharge Q is here represented by C.
For a rod of circular cross-section we have, taking the origin at the centre of the circle,
X = i(R 2 — x 2 — V 2 ), and the torsional rigidity is C = £/«r.R 4 . For the function tf> we have,
from (16.10), i// = constant. A constant xjt, however, corresponds by (16.4) to a simple dis-
placement of the whole rod along the s-axis, and so we can suppose that tfi = 0. Thus the
transverse sections of a circular rod undergoing torsion remain plane.
Problem 2. The same as Problem 1, but for an elliptical cross-section of semi-axes a
and b.
Solution. The torsional rigidity is C = 7r/xa s 6 s /(a s +6 2 ). The distribution of longitudinal
displacements is given by the torsion function ifi = (b* — a 2 )*y/(6*-f-a 2 ), where the co-ordinate
axes coincide with those of the ellipse.
Problem 3. The same as Problem 1, but for an equilateral triangular cross-section of
side a.
Solution. The torsional rigidity is C = \/3 /jui*J80. The torsion function is
i/j = y{xy/Z +y){x-y/3 —y)]6a
the origin being at the centre of the triangle and the x-axis along an altitude.
Problem 4. The same as Problem 1, but for a rod in the form of a long thin plate (of
width d and thickness h<^d).
Solution. The problem is equivalent to that of viscous fluid flow between plane parallel
walls. The result is that C = ^fidh s .
Problem 5. The same as Problem 1, but for a cylindrical pipe of internal and external
radii i? x and R 2 respectively.
Solution. The function x = i(-^2 a — r 2 ) (in polar co-ordinates) satisfies the condition
(16.13) at both boundaries of the annular cross-section of the pipe. From formula (16.17)
we then find C = i/wKiV — i?i 4 ).
Problem 6. The same as Problem 1, but for a thin- walled pipe of arbitrary cross-section.
Solution. Since the walls are thin, we can assume that x varies through the wall thickness
h, from zero on one side to Xi on the other, according to the linear law x = XiVlh (y being a
§17 Bending of rods 75
co-ordinate measured through the wall). Then the condition (16.13) gives XiUh = S,
where L is the perimeter of the pipe cross-section and S the area which it encloses. The
second term in the expression (16.17) is small compared with the first, and we obtain
C = 4hS 2 ^IL. If the pipe is cut longitudinally along a generator, the torsional rigidity falls
sharply, becoming (by the result of Problem 4) C = \ixLtf.
§17. Bending of rods
A bent rod is stretched at some points and compressed at others. Lines on
the convex side of the bent rod are extended, and those on the concave side
are compressed. As with plates, there is a neutral surface in the rod, which
undergoes neither extension nor compression. It separates the region of
compression from the region of extension.
Let us begin by investigating a bending deformation in a small portion of
the length of the rod, where the bending may be supposed slight; by this we
here mean that not only the strain tensor but also the magnitudes of the dis-
placements of points in the rod are small. We take a co-ordinate system with
the origin on the neutral surface in the portion considered, and the sr-axis
parallel to the axis of the undeformed rod. Let the bending occur in the
•sw-plane.f
As in the bending of plates and the twisting of rods, the external forces on
the sides of a thin bent rod are small compared with the internal stresses, and
can be taken as zero in determining the boundary conditions at the sides of the
rod. Thus we have everywhere on the sides of the rod a^njc = 0, or, since
n z = 0, a xx n x + a xy n y = 0, and similarly for i = y, z. We take a point on
the circumference of a cross-section for which the normal n is parallel to the
ar-axis. There will be another such point somewhere on the opposite side
of the rod. At both these points % = 0, and the above equation gives
a xx = 0. Since the rod is thin, however, a xx must be small everywhere in the
cross-section if it vanishes on either side. We can therefore put a xx =
everywhere in the rod. In a similar manner, it can be seen that all the com-
ponents of the stress tensor except a zz must be zero. That is, in the bending
of a thin rod only the extension (or compression) component of the internal
stress tensor is large. A deformation in which only the component a zz of
the stress tensor is non-zero is just a simple extension or compression (§5).
Thus there is a simple extension or compression in every volume element of
a bent rod. The amount of this varies, of course, from point to point in every
cross-section, and so the whole rod is bent.
It is easy to determine the relative extension at any point in the rod. Let
us consider an element of length dz parallel to the axis of the rod and near
the origin. On bending, the length of this element becomes dz'. The only
elements which remain unchanged are those which lie in the neutral surface.
Let R be the radius of curvature of the neutral surface near the origin. The
f In a rod undergoing only small deflections we can suppose that the bending occurs in a single
plane. This follows from the result of differential geometry that the deviation of a slightly bent curve
from a plane (its torsion) is of a higher order of smallness than its curvature.
76 The Equilibrium of Rods and Plates §17
lengths dz and dz' can be regarded as elements of arcs of circles whose radii
are respectively R and R + x, x being the co-ordinate of the point where
dz f lies. Hence
, , R + x , / x\
The relative extension is therefore (dz' — dz)/dz = xjR.
The relative extension of the element dz, however, is equal to the com-
ponent u zz of the strain tensor. Thus
u zz = x/R. (17.1)
We can now find a zz by using the relation a zz - Eu zz which holds for a
simple extension. This gives
a zz = ExjR. (17.2)
The position of the neutral surface in a bent rod has now to be determined.
This can be done from the condition that the deformation considered must
be pure bending, with no general extension or compression of the rod. The
total internal stress force on a cross-section of the rod must therefore be
zero, i.e. the integral J a zz df, taken over a cross-section, must vanish. Using
the expression (17.2) for a zz , we obtain the condition
jxdf=0. (17.3)
We can now bring in the centre of mass of the cross-section, which is that
of a uniform flat disc of the same shape. The co-ordinates of the centre of
mass are, as we know, given by the integrals J x d/// df, $y d// / df. Thus the
condition (17.3) signifies that, in a co-ordinate system with the origin in the
neutral surface, the x co-ordinate of the centre of mass of any cross-section
is zero. The neutral surface therefore passes through the centres of mass
of the cross-sections of the rod.
Two components of the strain tensor besides u zz are non-zero, since for a
simple extension we have u X x = tt yy = —au zz . Knowing the strain tensor,
we can easily find the displacement also :
u zz — du z Jdz = xjR, duzjdx = du y /dy = — ax/R,
du z du x du x du v du u du z
— -+ — - = 0, +^=0, — - + — ~ = 0.
dx dz By dx dz dy
Integration of these equations gives the following expressions for the com-
ponents of the displacement :
"^"a^ 2 "^ 2- ^' (17.4)
u y = — axyJR, u z = xz/R.
The constants of integration have been put equal to zero; this means that
we "fix" the origin.
§17
Bending of rods
77
It is seen from formulae (17.4) that the points initially on a cross-section
z — constant = %o will be found, after the deformation, on the surface
z = zo+u z = zo(l+xlR). We see that, in the approximation used, the
cross-sections remain plane but are turned through an angle relative to their
initial positions. The shape of the cross-section changes, however; for
example, when a rod of rectangular cross-section (sides a, b) is bent, the sides
y = + \b of the cross-section become y = ±\b + u y = ±\b{\ — ax\R\ i.e.
no longer parallel but still straight. The sides x = ±\a> however, are bent
into the parabolic curves
(Fig. 14).
x = ±\a + u x = ±\a- — [*o 2 + <Ki« 2 -:>' 2 )]
Lis.
— j— b — •-
n
a
\
1
/
1
1
i
\
\
Fig. 14
The free energy per unit volume of the rod is
\<JikUQz = \o lz u zz = %Ex 2 /R 2 .
Integrating over the cross-section of the rod, we have
K£/* 2 )J** 2 d/.
(17.5)
This is the free energy per unit length of a bent rod. The radius of curvature
R is that of the neutral surface. However, since the rod is thin, JR can here
be regarded, to the same approximation, as the radius of curvature of the
bent rod itself, regarded as a line (often called an "elastic line").
In the expression (17.5) it is convenient to introduce the moment of
inertia of the cross-section. The moment of inertia about they-axis in its plane
is defined as
/„« J>d/,
(17.6)
analogously to the ordinary moment of inertia, but with the surface element
d/ instead of the mass element. Then the free energy per unit length of the
rod can be written
\EI y IR\
(17.7)
78 The Equilibrium of Rods and Plates §18
We can also determine the moment of the internal stress forces on a given
cross-section of the rod (the bending moment). A force a zz d/ = (xE/R) d/
acts in the z-direction on the surface element d/ of the cross-section. Its
moment about the y-axis is xa zz d/. Hence the total moment of the forces
about this axis is
My = (E/R) jx*df= ElyfR. (17.8)
Thus the curvature \jR of the elastic line is proportional to the bending
moment on the cross-section concerned.
The magnitude of I y depends on the direction of the jy-axis in the cross-
sectional plane. It is convenient to express I y in terms of the principal
moments of inertia. If 6 is the angle between the jy-axis and one of the
principal axes of inertia in the cross-section, we know from mechanics that
I y = h cos 2 d+I 2 sin 2 0, (17.9)
where h and 7 2 are the principal moments of inertia. The planes through
the #-axis and the principal axes of inertia are called the principal planes of
bending.
If, for example, the cross-section is rectangular (with sides a, b), its centre
of mass is at the centre of the rectangle, and the principal axes of inertia
are parallel to the sides. The principal moments of inertia are
h = a*b/12, h = 0& 3 /12. (17.10)
For a circular cross-section of radius R, the centre of mass is at the centre
of the circle, and the principal axes are arbitrary. The moment of inertia
about any axis lying in the cross-section and passing through the centre is
/ = IttR\ (17.11)
§18. The energy of a deformed rod
In §17 we have discussed only a small portion of the length of a bent rod.
In going on to investigate the deformation throughout the rod, we must
begin by finding a suitable method of describing this deformation. It is
important to note that, when a rod undergoes large bending deflections,-)-
there is in general a twisting of it as well, so that the resulting deformation
is a combination of pure bending and torsion.
To describe the deformation, it is convenient to proceed as follows. We
divide the rod into infinitesimal elements, each of which is bounded by two
adjacent cross-sections. For each such element we use a co-ordinate system
f , rj, £, so chosen that all the systems are parallel in the undeformed state,
and their £-axes are parallel to the axis of the rod. When the rod is bent, the
t By this, it should be remembered, we mean that the vector u is not small, but the strain tensor
is still small.
§18 The energy of a deformed rod 79
co-ordinate system in each element is rotated, and in general differently in
different elements. Any two adjacent systems are rotated through an infini-
tesimal relative angle.
Let d<|> be the vector of the angle of relative rotation of two systems at a
distance d/ apart along the rod (we know that an infinitesimal angle of rotation
can be regarded as a vector parallel to the axis of rotation ; its components
are the angles of rotation about each of the three axes of co-ordinates).
To describe the deformation, we use the vector
SI = d*/d/, (18.1)
which gives the "rate" of rotation of the co-ordinate axes along the rod. If
the deformation is a pure torsion, the co-ordinate system rotates only about
the axis of the rod, i.e. about the £-axis. In this case, therefore, the vector
SI is parallel to the axis of the rod, and is just the torsion angle r used in §16.
Correspondingly, in the general case of an arbitrary deformation we can call
the component Q^ of the vector SI the torsion angle. For a pure bending of the
rod in a single plane, on the other hand, the vector SI has no component Q^,
i.e. it lies in the l^-plane at each point. If we take the plane of bending as the
££-plane, then the rotation is about the ry-axis at every point, i.e. SI is parallel
to the 77-axis.
We take a unit vector t tangential to the rod (regarded as an elastic line).
The derivative dt/d/ is the curvature vector of the line; its magnitude is
1/R, where R is the radius of curvature, f and its direction is that of the
principal normal to the curve. The change in a vector due to an infinitesimal
rotation is equal to the vector product of the rotation vector and the vector
itself. Hence the change in the vector t between two neighbouring points of
the elastic line is given by dt = d<t> X t, or, dividing by d/,
dt/d/= Slxt. (18.2)
Multiplying this equation vectorially by t, we have
SI = txdt/d/+t(t- a). (18.3)
The direction of the tangent vector at any point is the same as that of the
£-axis at that point. Hence t • SI = Q f . Using the unit vector n along the
principal normal (n = R dt/d/), we can therefore put
SI = tXnIR + tfy. (18.4)
The first term on the right is a vector with two components Q g , £l r
The unit vector t Xn is the binormal unit vector. Thus the components Q £ ,
Q v form a vector along the binormal to the rod, whose magnitude equals the
curvature 1/R.
f It may be recalled that any curve in space is characterised at each point by a curvature and a
torsion. This torsion (which we shall not use) should not be confused with the torsional deformation,
which is a twisting of a rod about its axis.
80 The Equilibrium of Rods and Plates §18
By using the vector SI to characterise the deformation and ascertaining
its properties, we can derive an expression for the elastic free energy of a
bent rod. The elastic energy per unit length of the rod is a quadratic function
of the deformation, i.e., in this case, a quadratic function of the components
of the vector SI. It is easy to see that there can be no terms in this quadratic
form proportional to QgQg and Q^Q^. For, since the rod is uniform along
its length, all quantities, and in particular the energy, must remain constant
when the direction of the positive £-axis is reversed, i.e. when £ is replaced
by — £, whereas the products mentioned change sign.
For Qg = Q = we have a pure torsion, and the expression for the
energy must be that obtained in §16. Thus the term in Q^ 2 in the free
energy is \C£l^.
Finally, the terms quadratic in Q.g and Q. v can be obtained by starting from
the expression (17.7) for the energy of a slightly bent short section of the rod.
Let us suppose that the rod is only slightly bent. We take the ££-plane as
the plane of bending, so that the component Qg is zero ; there is also no torsion
in a slight bending. The expression for the energy must then be that given
by (17.7), i.e. %EI v jR 2 . We have seen, however, that 1/JR 2 is the square of the
two-dimensional vector (Q.%, Q. v ). Hence the energy must be of the foi'm
^ElyQ,^. For an arbitrary choice of the £ and v\ axes this expression becomes,
as we know from mechanics,
i£(/„Q f » + 2Z,gQ ? Q £ +/ K £y),
where I V7]y / g, 1^ are the components of the inertia tensor for the cross-
section of the rod. It is convenient to take the £ and 17 axes to coincide with
the principal axes of inertia. We then have simply ^E(IiQ^ 2 + l2Sl v 2 ) } where
7i, I2 are the principal moments of inertia. Since the coefficients of £2g 2 and
Q, v 2 are constants, the resulting expression must be valid for large deflections
also.
Finally, integrating over the length of the rod, we obtain the following
expression for the elastic free energy of a bent rod :
Frod = J* {ihEQf + lhEnz + iCn&dl. (18.5)
Next, we can express in terms of SI the moment of the forces acting on
a cross-section of the rod. This is easily done by again using the results
previously obtained for pure torsion and pure bending. In pure torsion, the
moment of the forces about the axis of the rod is Cr. Hence we conclude
that, in the general case, the moment M^ about the £-axis must be CD.^.
Next, in a slight deflection in the f£-plane, the moment about the 77-axis is
EhjR. In such a bending, however, the vector SI is along the 17-axis, so that
1/R is just the magnitude of SI, and EI 2 /R = EI 2 Q. Hence we conclude
that, in the general case, we must have M g — EIi£l g , M v = EI 2 & V (the $ and
r\ axes being along the principal axes of inertia in the cross-section). Thus
Frod = J"|
§18 The energy of a deformed rod 81
the components of the moment vector M are
M { = Em gy M v = Eh& r M c = CQ C . (18.6)
The elastic energy (18.5), expressed in terms of the moment of the forces, is
Mr 2 M 2 M? \
i^!_ + i!!i_ + ^i- d/. (18.7)
ZhE 2hE 2C J
An important case of the bending of rods is that of a slight bending, in
which the deviation from the initial position is everywhere small compared
with the length of the rod. In this case torsion can be supposed absent, and
we can put Q £ = 0, so that (18.4) gives simply
Sl = txn//? = txdt/d/. (18.8)
We take a co-ordinate system x, y, z fixed in space, with the 2-axis along the
axis of the undeformed rod (instead of the system £, rj, £ for each point in the
rod), and denote by X, Y the co-ordinates x, y for points on the elastic line;
X and Y give the displacement of points on the line from their positions
before the deformation.
Since the bending is only slight, the tangent vector t is almost parallel
to the .sr-axis, and the difference in direction can be approximately neglected.
The unit tangent vector is the derivative t = dr/d/ of the radius vector r
of a point on the curve with respect to its length. Hence
dt/d/= d 2 r/d/ 2 ^d 2 r/d* 2 ;
the derivative with respect to the length can be approximately replaced by
the derivative with respect to z. In particular, the x and y components of
this vector are respectively d 2 Xjdz 2 and d 2 Y/d;s 2 . The components ft £ , Cl v
are, to the same accuracy, equal to Q xt Q v , and we have from (18.8)
Q g = -d 2 Y/d* 2 , Q v = d 2 X/d* 2 . (18.9)
Substituting these expressions in (18.5), we obtain the elastic energy of a
slightly bent rod in the form
ri /d 2 Y\ 2 /d*X\ 2 \
F <«^ E iH-w) +h (-^)h (18 - 10)
Here h and h are the moments of inertia about the axes of x and y respectively,
which are the principal axes of inertia.
In particular, for a rod of circular cross-section, h = h = A a nd the
integrand is just the sum of the squared second derivatives, which in the
approximation considered is the square of the curvature :
/d*x\ 2 /d 2 yy
\ow + \d^7
/d*X\ 2 /d 2 Y\ 2 1
82 The Equilibrium of Rods and Plates §19
Hence formula (18.10) can be plausibly generalised to the case of slight
bending of a circular rod having any shape (not necessarily straight) in its
undeformed state. To do so, we must write the bending energy as
Frod = \EI J(— - — J d*, (18.11)
where R is the radius of curvature at any point of the undeformed rod. This
expression has a minimum, as it should, in the undeformed state (R = R ),
and for R -> co it becomes formula (18.10).
§19. The equations of equilibrium of rods
We can now derive the equations of equilibrium for a bent rod. We again
consider an infinitesimal element bounded by two adjoining cross-sections
of the rod, and calculate the total force acting on it. We denote by F the
resultant internal stress on a cross-section.f The components of this vector
are the integrals of a^ over the cross-section :
Fi^ja^df. (19.1)
If we regard the two adjoining cross-sections as the ends of the element, a
force F+dF acts on the upper end, and -F on the lower end; the sum of
these is the differential dF. Next, let K be the external force on the rod per
unit length. Then an external force K dl acts on the element of length d/.
The resultant of the forces on the element is therefore dF+K d/. This must
be zero in equilibrium. Thus we have
dF/dZ = - K. (19.2)
A second equation is obtained from the condition that the total moment of
the forces on the element is zero. Let M be the moment of the internal
stresses on the cross-section. This is the moment about a point (the origin)
which lies in the plane of the cross-section; its components are given by
formulae (18.6). We shall calculate the total moment, on the element con-
sidered, about a point O lying in the plane of its upper end. Then the
internal stresses on this end give a moment M+dM. The moment about O
of the internal stresses on the lower end of the element is composed of the
moment — M of those forces about the origin O' in the plane of the lower
end and the moment about O of the total force — F on that end. This latter
moment is — dl X — F, where dl is the vector of the element of length of the
rod between O' and O. The moment due to the external forces K is of a
higher order of smallness. Thus the total moment acting on the element
considered is dM + dlxF. In equilibrium, this must be zero:
dM + dlxF = 0.
| This notation will not lead to any confusion with the free energy, which does not appear in
§§19-21.
§19 The equations of equilibrium of rods 83
Dividing this equation by d/ and using the fact that dl/d/ = t is the unit
vector tangential to the rod (regarded as a line), we have
dM/d/=Fxt. (19.3)
Equations (19.2) and (19.3) form a complete set of equilibrium equations
for a rod bent in any manner.
If the external forces on the rod are concentrated, i.e. applied only at
isolated points of the rod, the equilibrium equations at all other points are
much simplified. For K = we have from (19.2)
F = constant, (19.4)
i.e. the stress resultant is constant along any portion of the rod between
points where forces are applied. The values of the constant are found from
the fact that the difference F 2 -Fi of the forces at two points 1 and 2 is
F 2 -Fi= -SK, (19.5)
where the sum is over all forces applied to the segment of the rod between
the two points. It should be noticed that, in the difference F2-F1, the
point 2 is further from the point from which / is measured than is the point 1 ;
this is important in determining the signs in equation (19.5). In particular, if
only one concentrated force f acts on the rod, and is applied at its free end,
then F = constant = f at all points of the rod.
The second equilibrium equation (19.3) is also simplified. Putting
t = dl/d/ = dr/d/ (where r is the radius vector from any fixed point to the
point considered) and integrating, we obtain
M = FXr+ constant, (19.6)
since F is constant.
If concentrated forces also are absent, and the rod is bent by the application
of concentrated moments, i.e. of concentrated couples, then F = constant
at all points of the rod, while M is discontinuous at points where couples
are applied, the discontinuity being equal to the moment of the couple.
Let us consider also the boundary conditions at the ends of a bent rod.
Various cases are possible.
The end of the rod is said to be clamped (Fig. 4a, §12) if it cannot move
either longitudinally or transversely, and moreover its direction (i.e. the direc-
tion of the tangent to the rod) cannot change. In this case the boundary
conditions are that the co-ordinates of the end of the rod and the unit tangen-
tial vector t there are given. The reaction force and moment exerted on the
rod by the clamp are determined by solving the equations.
The opposite case is that of a free end, whose position and direction are
arbitrary. In this case the boundary conditions are that the force F and
moment M must be zero at the end of the rod.f
t If a concentrated force f is applied to the free end of the rod, the boundary condition is F = f
not F = 0.
84 The Equilibrium of Rods and Plates §19
If the end of the rod is fixed to a hinge, it cannot be displaced, but its
direction can vary. In this case the moment of the forces on the freely turning
end must be zero.
Finally, if the rod is supported (Fig. 4b), it can slide at the point of support
but cannot undergo transverse displacements. In this case the direction t of
the rod at the support and the point on the rod at which it is supported are
unknown. The moment of the forces at the point of support must be zero,
since the rod can turn freely, and the force F at that point must be perpen-
dicular to the rod; a longitudinal force would cause a further sliding of the
rod at this point.
The boundary conditions for other modes of fixing the rod can easily be
established in a similar manner. We shall not pause to add to the typical
examples already given.
It was mentioned at the beginning of §18 that a rod of arbitrary cross-
section undergoing large deflections is in general twisted also, even if no
external twisting moment is applied to the rod. An exception occurs when a
rod is bent in one of its principal planes, in which case there is no torsion.
For a rod of circular cross-section no torsion results for any bending (if there
is no external twisting moment, of course). This can be seen as follows. The
twisting is given by the component Q% = SI • t of the vector SI. Let us
calculate the derivative of this along the rod. To do so, we use the fact that
Q c = MJC:
d „, x ~ d ^ dM dt
— (M-t) = C — *- = t+M — .
d/ v ' 61 61 61
Substituting (19.3), we see that the first term is zero, so that
C6SiJ61= M-dt/d/.
For a rod of circular cross-section, h = I 2 = I; by (18.3) and (18.6), we can
therefore write M in the form
M = EItx6t/6l+tCQ c . (19.7)
Multiplying by dt/d/, we have zero on the right-hand side, so that
6QJ61 = 0,
whence
D s = constant, (19.8)
i.e. the torsion angle is constant along the rod. If no twisting moments are
applied to the ends of the rod, then Q,^ is zero at the ends, and there is no
torsion anywhere in the rod.
For a rod of circular cross-section, we can therefore put for pure bending
dr d 2 r
M = EIX Xdt/61 = EI— x— . (19.9)
61 d/ 2
§19 The equations of equilibrium of rods 85
Substituting this in (19.3), we obtain the equation for pure bending of a
circular rod:
<fr d 3 r dr ,« rt < A v
£/ d7*d^ = Fx d? (mo)
PROBLEMS
Problem 1. Reduce to quadratures the problem of determining the shape of a rod of
circular cross-section bent in one plane by concentrated forces.
Solution. Let us consider a portion of the rod lying between points where the forces
are applied ; on such a portion F is constant. We take the plane of the bent rod as the xy-
plane, with the y-axis parallel to the force F, and introduce the angle 8 between the tangent
to the rod and the y-axis. Then dxjdl = sin 8, dyjdl = cos 8, where x, y are the co-ordinates
of a point on the rod. Expanding the vector products in (19.10), we obtain the following
equation for 8 as a function of the arc length I: £7d 2 0/d/ a — Fsin 8 = 0. A first integration
gives lEI(dOld[)*+Fcos 8 = c lt and
/ = ± VdEI) f ,. Z ^+C2. (1)
J v(ci — Fcosv)
The function 8(1) can be obtained in terms of elliptic functions. The co-ordinates
x = J sin 6 dl t y = j cos 6 d/
are
x = ± y/[2EI(ci - F cos 6)1 F 2 ] + constant,
r cos0d0 (2)
y = +W(iEI) + constant.
The moment M (19.9) is parallel to the ar-axis, and its magnitude is M = EId8/dl.
Problem 2. Determine the shape of a bent rod with one end clamped and the other under
a force f perpendicular to the original direction of the rod (Fig. 15).
Fig. 15
Solution. We have F = constant = f everywhere on the rod. At the clamped end
(/ = 0), 8 = \it, and at the free end (Z = L, the length of the rod) M = 0, i.e. 8' = 0. Putting
8(L) = 8 , we have in (1), Problem 1, Ci — /cos 8 and
= VQEiiflj
dO
-v/(cos do — cos 8)
86 The Equilibrium of Rods and Plates §19
Hence we obtain the equation for 6 :
= V(iEIlf)j
dd
\/(cos #o — cos 6)
The shape of the rod is given by
x = ^/ (2EI jf)[^/ (cos 6 ) -V (cos do -cos 6) \,
in
y = V(EI/2f)j
cosddd
yYcos do — cos 6)
e
Problem 3. The same as Problem 2, but for a force f parallel to the original direction of
the rod.
Fig. 16
Solution. We have F = — f; the co-ordinate axes are taken as shown in Fig. 16. The
boundary conditions are 8 = for I = 0, 0' = for / = L. Then
vwwj
ie
yYcos 6 — cos #o)
o
where
d = 6(L)
is given by
0.
- vmif)j
dd
\/(cOS0— COS0o)
For * and y we obtain
x = V(2W)[V(l~ cos 0o)- V(cos0-cos0 o )],
e
y=V(EI/2f)j
cos 6 dd
-\/(cos 6 — cos do)
§19
The equations of equilibrium of rods
87
For a small deflection, O <^ 1, and we can write
L
v(£///) !v(^)= w(£///) '
i.e. 0q does not appear. This shows that, in accordance with the result of §21, Problem 3,
the solution in question exists only for / 5s ir 2 EI14L 2 , i.e. when the rectilinear shape ceases
to be stable.
Problem 4. The same as Problem 2, but for the case where both ends of the rod are sup-
ported and a force f is applied at its centre. The distance between the supports is L .
Solution. We take the co-ordinate axes as shown in Fig. 17. The force F is constant on
each of the segments AB and BC, and on each is perpendicular to the direction of the rod at
the point of support A or C. The difference between the values of F on AB and BC is f,
and so we conclude that, on AB, F sin 8 = — i/, where o is the angle between the y-axis
and the line AC. At the point A (/ = 0) we have the conditions = \tt and M = 0, i.e.
0' = 0, so that on AB
I EI sin O r
= V ? Ji
/
■\/cos8
x = 2
EI sin #o cos 6
1 '
y =
/EI sin 6 f
cos0d0.
The angle 8 is determined from the condition that the projection of AB on the straight line
AC must be iL , whence
\u
J EI sin O fi
= v7 J"
EI sin O fcos(0-0 o )
<\/sin0
d0.
*•
For some value O lying between and Jw the derivative dfld0 o (/being regarded as a function
of 6 ) passes through zero to positive values. A further decrease in O > i-e. increase in the
deflection, would mean a decrease in /. This means that the solution found here becomes
unstable, the rod collapsing between the supports.
Problem 5. Reduce to quadratures the problem of three-dimensional bending of a rod
under the action of concentrated forces.
88 The Equilibrium of Rods and Plates §19
Solution. Let us consider a segment of the rod between points where forces are applied,
on which F = constant. Integrating (19.10), we obtain
dr d2r
the constant of integration has been written as a vector cF parallel to F, since, by appro-
priately choosing the origin, i.e. by adding a constant vector to r, we can eliminate any vector
perpendicular to F. Multiplying (1) scalar ly and vectorially by r' (the prime denoting
differentiation with respect to I), and using the fact that r'«r" = (since r' 2 = 1), we obtain
FTXr'+cF^r' = 0, EIv" = (Fxr)Xr'+cFxr'. In components (with the ar-axis parallel
to F) we obtain (xy' — yx') + cz' = 0, EIz" = —F(xx'+yy'). Using cylindrical polar co-
ordinates r, <f>, z, we have
r*<f>' + cz' = 0, EIz" = -Err'. (2)
The second of these gives
z' = F{A-r*)\2Eh (3)
where A is a constant. Combining (2) and (3) with the identity r' 2 +r*^'* -J-*'* = 1, we find
r dr
d/ =
VI> 2 - (r 2 + c z )(A - r2)2F2/4£2/2j'
and then (2) and (3) give
{A-r*)rdr
z =
/
c
2EI J y/[r* - F 2 (r* + c*)(A - r^ffim*]
(A-r*)dr
V[r 2 - F 2 (r 2 + c 2 )(A - r2)*/4£ 2 / 2 ]
which gives the shape of the bent rod.
Problem 6. A rod of circular cross-section is subjected to torsion (with torsion angle r)
and twisted into a spiral. Determine the force and moment which must be applied to the
ends of the rod to keep it in this state.
Solution. Let R be the radius of the cylinder on whose surface the spiral lies (and along
whose axis we take the ^-direction) and a the angle between the tangent to the spiral and a
plane perpendicular to the sr-axis ; the pitch h of the spiral is related to a and R by h — 2nR tan a.
The equation of the spiral is x — R cos <$>, y = R sin <f>, z = <j>R tan a, where <f> is the angle
of rotation about the »-axis. The element of length is 61 — (Rjcos a)d<£. Substituting these
expressions in (19.7), we calculate the components of the vector M, and then the force F
from formula (19.3); F is constant everywhere on the rod. The result is that the force F is
parallel to the z-axis and its magnitude is F = F z = (O/ R) sin a— (EI/R 3 ) cos 2 a sin a.
The moment M has a ^-component M t — Cr sin a-\-(EI/R) cos 3 a and a ^-component, along
the tangent to the cross-section of the cylinder, M$ = FR.
Problem 7. Determine the form of a flexible wire (whose resistance to bending can be
neglected in comparison with its resistance to stretching) suspended at two points and in a
gravitational field.
Solution. We take the plane of the wire as the jry-plane, with the y-axis vertically down-
wards. In equation (19.3) we can neglect the term dM/dZ, since M is proportional to EI.
Then Fxt = 0, i.e. F is parallel to t at every point, and we can put F = Ft. Equation (19.2)
then gives
d / dx\ n d ( n dy\
d7
A dl! dl\ 61 J H
§20 Small deflections of rods 89
where q is the weight of the wire per unit length; hence F 6x1 61 = c, Fdyldl — ql, and so
F = V(«"+flV). so that dxldl = A/V(A 2 +P), dyjdl = UV(A*+P), where ^4 = elq.
Integration gives x = A sivh-\llA), y — ^(A*+P), whence y = A cosh (x/A), i.e. the
wire takes the form of a catenary. The choice of origin and the constant A are determined
by the fact that the curve must pass through the two given points and have a given length.
§20. Small deflections of rods
The equations of equilibrium are considerably simplified in the important
case of small deflections of rods. This case holds if the direction of the
vector t tangential to the rod varies only slowly along its length, i.e. the deriva-
tive dt/d/ is small. In other words, the radius of curvature of the bent rod is
everywhere large compared with the length of the rod. In practice, this
condition amounts to requiring that the transverse deflection of the rod is
small compared with its length. It should be emphasised that the deflection
need not be small compared with the thickness of the rod, as it had to be in
the approximate theory of small deflections of plates given in §§ll-12.f
Differentiating (19.3) with respect to the length, we have
d2M dF dt ^^
= Xt+Fx— . (20.1)
d/2 61 dl K '
The second term contains the small quantity dt/d/, and so can usually be
neglected (some exceptional cases are discussed below). Substituting in the
first term dF/d/ = — K, we obtain the equation of equilibrium in the form
d 2 M/d/2 = txK. (20.2)
We write this equation in components, substituting in it from (18.6) and
(18.9)
M z = -EhY", My = EIzX", M z = 0, (20.3)
where the prime denotes differentiation with respect to z. The unit vector t
may be supposed to be parallel to the #-axis. Then (20.2) gives
EI 2 X™-K X = 0, EhY^-Ky = 0. (20.4)
These equations give the deflections X and Y as functions of z, i.e. the shape
of a slightly bent rod.
The stress resultant F on a cross-section of the rod can also be expressed in
terms of the derivatives of X and Y. Substituting (20.3) in (19.3), we obtain
F x = -EhX"\ F y = -EhY"'. (20.5)
We see that the second derivatives give the moment of the internal stresses,
while the third derivatives give the stress resultant. The force (20.5) is
called the shearing force. If the bending is due to concentrated forces, the
shearing force is constant along each segment of the rod between points
f We shall not give the complex theory of the bending of rods which are not straight when un-
deformed, but only consider one simple example (see Problems 8 and 9).
90 The Equilibrium of Rods and Plates §20
where forces are applied, and has a discontinuity at each of these points
equal to the force applied there.
The quantities EI% and EI\ are called the flexural rigidities of the rod in
the xz and yz planes respectively, f
If the external forces applied to the rod act in one plane, the bending takes
place in one plane, though not in general the same plane. The angle between
the two planes is easily found. If a is the angle between the plane of action of
the forces and the first principal plane of bending (the a:2-plane), the equa-
tions of equilibrium become X* iv > = (K/I 2 E) cos a, Y< lv > = (KlhE) sin a.
The two equations differ only in the coefficient of K. Hence X and Y are
proportional, and Y = (Xh/Ii) tan a. The angle 6 between the plane of
bending and the xsr-plane is given by
tan0 = (7 2 //i)tana. (20.6)
For a rod of circular cross-section h = h and <x = 6, i.e. the bending occurs
in the plane of action of the forces. The same is true for a rod of any cross-
section when a = 0, i.e. when the forces act in a principal plane. The magni-
tude of the deflection £ = V( X2 + Y2 ) satisfies the equation
£/£(iv) = k, I = IihlVVi 2 cos2 a + / 2 2 sin2 a ). (20.7)
The shearing force F is in the same plane as K, and its magnitude is
F = -Ell'". (20.8)
Here I is the "effective" moment of inertia of the cross-section of the rod.
We can write down explicitly the boundary conditions on the equations of
equilibrium for a slightly bent rod. If the end of the rod is clamped, we must
have X = Y = there, and also X' = Y' = 0, since its direction cannot
change. Thus the conditions at a clamped end are
X = Y = 0, X' = y = 0. (20.9)
The reaction force and moment at the point of support are determined from
the known solution by formulae (20.3) and (20.5).
When the bending is sufficiently slight, the hinging and supporting of a
point on the rod are equivalent as regards the boundary conditions. The
reason is that, in the latter case, the longitudinal displacement of the rod at
its point of support is of the second order of smallness compared with the
f An equation of the form
DXW-K X = (20.4a)
also describes the bending of a thin plate in certain limiting cases. Let a rectangular plate (with
sides a, b and thickness h) be fixed along its sides a (parallel to the y-axis) and bent along its sides b
(parallel to the 2-axis) by a load uniform in the y-direction. In the general case of arbitrary a and b,
the two-dimensional equation (12.5), with the appropriate boundary conditions at the fixed and free
edges, must be used to determine the bending. In the limiting case a g> b, however, the deformation
may be regarded as uniform in the y-direction, and then the two-dimensional equilibrium equation
becomes of the form (20.4a), with the flexural rigidity replaced by D = E/i 3 a/12(l -o 2 ). Equation
(20.4a) is also applicable to the opposite limiting case a <g b, when the plate can be regarded as a
rod of length b with a narrow rectangular cross-section (a rectangle of sides a and h); in this case,
however, the flexural rigidity is D = EI2 = Eh 3 ajl2.
§20 Small deflections of rods 91
transverse deflection, and can therefore be neglected. The boundary con-
ditions of zero transverse displacement and moment give
X = Y = 0, X" = Y" = 0. (20.10)
The direction of the end of the rod and the reaction force at the point of
support are obtained by solving the equations.
Finally, at a free end, the force F and moment M must be zero. According
to (20.3) and (20.5), this gives the conditions
X" = Y" = 0, X'" = Y'" = 0. (20.11)
If a concentrated force is applied at the free end, then F must be equal to this
force, and not to zero.
It is not difficult to generalise equations (20.4) to the case of a rod of
variable cross-section. For such a rod the moments of inertia I\ and 1% are
functions of z. Formulae (20.3), which determine the moment at any cross-
section, are still valid. Substitution in (20.2) now gives
d 2 / d 2 Y\ d 2 / d 2 ^\
E — (h = K y , E—(h — A = K *> ( 20A2 )
d*2\ dW d* 2 \ <W
in which Ij and h must be differentiated. The shearing force is
d / d 2 X\ d / d 2 Y\
' — \h , F y = -E — (h . (20.13)
d*\ d* 2 / V d*\ d* 2 / v '
Let us return to equations (20.1). Our neglect of the second term on the
right-hand side may in some cases be illegitimate, even if the bending is
slight. The cases involved are those in which a large internal stress resultant
acts along the rod, i.e. F z is very large. Such a force is usually caused by a
strong tension of the rod by external stretching forces applied to its ends.
We denote by T the constant lengthwise stress F z . If the rod is strongly
compressed instead of being extended, T will be negative. In expanding the
vector product F Xdt/d/ we must now retain the terms in T, but those in F x
and F y can again be neglected. Substituting X", Y", 1 for the components
of the vector dt/d/, we obtain the equations of equilibrium in the form
hEX^)-TX"-K x = 0,
(20.14)
I 1 EYM-TY"-K v = 0. v '
The expressions (20.5) for the shearing force will now contain additional
terms giving the projections of the force T (along the vector t) on the x and
y axes :
F x = -EI 2 X'" + TX', F y = -EhY'" + TY'. (20.15)
These formulae can also, of course, be obtained directly from (19.3).
In some cases a large force T can result from the bending itself, even if
no stretching forces are applied. Let us consider a rod with both ends
F x = -E
92 The Equilibrium of Rods and Plates §20
clamped or hinged to fixed supports, so that no longitudinal displacement is
possible. Then the bending of the rod must result in an extension of it,
which leads to a force T in the rod. It is easy to estimate the magnitude of the
deflection for which this force becomes important. The length L + AL of the
bent rod is given by
L
L+AL = JV(l + *' 2 + Y' 2 ) d*,
o
taken along the straight line joining the points of support. For slight bending
the square root can be expanded in series, and we find
L
The stress force in simple stretching is equal to the relative extension multi-
plied by Young's modulus and by the area S of the cross-section of the rod.
Thus the force T is
E'er L
T = — J(Z' 2 + Y'2) d*. (20.16)
2L
If 8 is the order of magnitude of the transverse bending, the derivatives
X' and Y' are of the order of 8jL, so that the integral in (20.16) is of the
order of S 2 /L, and T ~ ES(8/L) 2 . The orders of magnitude of the first and
second terms in (20.14) are respectively EI8/L* and T8/L 2 ~ ES8 3 /L*. The
moment of inertia / is of the order of A 4 , and S ~ h 2 , where h is the thickness
of the rod. Substituting, we easily find that the first and second terms in
(20.14) are comparable in magnitude if 8 ~ h. Thus, when a rod with fixed
ends is bent, the equations of equilibrium can be used in the form (20.4) only
if the deflection is small in comparison with the thickness of the rod. If 8
is not small compared with h (but still, of course, small compared with L),
equations (20.14) must be used. The force T in these equations is not known
a priori. It must first be regarded as a parameter in the solution, and then
determined by formula (20.16) from the solution obtained; this gives the
relation between T and the bending forces applied to the rod.
The opposite limiting case is that where the resistance of the rod to bending
is small compared with its resistance to stretching, so that the first terms in
equations (20.14) can be neglected in comparison with the second terms.
Physically this case can be realized either by a very strong tension force T or
by a small value of EI, which can result from a small thickness h. Rods under
strong tension are called strings. In such cases the equations of equilibrium
are
TX" + K X = 0, TY" + K y = 0. (20.17)
§20 Small deflections of rods 93
The ends of the string are fixed, in the sense that their co-ordinates are given,
i.e.
X = Y = 0. (20.18)
The direction of the ends cannot be decided arbitrarily, but is given by the
solution of the equations.
In conclusion, we may show how the equations of equilibrium of a slightly
bent rod may be obtained from the variational principle, using the expression
(18.10) for the elastic energy:
F to a = lEJ{hY"2 + l2X"2}dz.
In equilibrium the sum of this energy and the potential energy due to the
external forces K acting on the rod must be a minimum, i.e. we must have
8F TO &-$(K x 8X+K y 8Y)<\z = 0, where the second term is the work done
by the external forces in an infinitesimal displacement of the rod. In varying
■Prod, we effect a repeated integration by parts:
l8JX"*dz = jX"8X"dz
= [X"8X']- jX'"8X'dz
= [X ,, 8X']-[X ,,, 8X]+ jX™8Xdz,
and similarly for the integral of Y" 2 . Collecting terms, we obtain
j [(Eh yav) _ K y )8 Y + (EIzX<M - K X )8X] d* +
+ EI 1 [(Y"8Y , -Y , "8Y)] + EI 2 [(X f '8X f -X , "SX)] = 0.
The integral gives the equilibrium equations (20.4), since the variations 8X
and 8 Y are arbitrary. The integrated terms give the boundary conditions on
these equations ; for example, at a free end the variations 8X, 8 Y, 8X', 8 Y'
are arbitrary, and the corresponding conditions (20.11) are obtained. Also,
the coefficients of 8X and 8Y in these terms give the expressions (20.5) for
the components of the shearing force, and those of 8X' and 8Y' give the
expressions (20.3) for the components of the bending moment.
Finally, the equations of equilibrium (20.14) in the presence of a tension
force T can be obtained by the same method if we include in the energy a
term TAL = \T\ (X' 2 + Y' 2 ) dsr, which is the work done by the force T over a
distance AL equal to the extension of the rod.
PROBLEMS
Problem 1. Determine the shape of a rod (of length /) bent by its own weight, for various
modes of support at the ends.
94 The Equilibrium of Rods and Plates §20
Solution. The required shape is given by a solution of the equation £(' v > = qlEI, where
q is the weight per unit length, with the appropriate boundary conditions at its ends, as shown
in the text. The following shapes and maximum displacements are obtained for various
modes of support at the ends of the rod. The origin is at one end of the rod in each case.
(a) Both ends clamped:
£ = qz\z-lf\2\EI y £(£/) = qPfiMEI.
(b) Both ends supported :
£ = qz(z3-2lz* + P)/24EI, £(£Z) = 5ql*/384EI.
(c) One end (z = I) clamped, the other supported:
£ = qz(2z* - 3Z* 2 + /3)/48£7, £(0-42/) = 0-0054#/£7.
(d) One end (z — 0) clamped, the other free :
£ = qz\z 2 -\lz+W)\2\EI, £(Z) = ql*J8EI.
Problem 2. Determine the shape of a rod bent by a force /applied to its mid-point.
Solution. We have £( lv) = everywhere except at z = \l. The boundary conditions
at the ends of the rod (z = and z — I) are determined by the mode of support ; at z = £/,
£, £' and £" must be continuous, and the discontinuity in the shearing force F = —Ell,'"
must be equal to/.
The shape of the rod (for < z ^ £/) and the maximum displacement are given by the
following formulae:
(a) Both ends clamped :
£ = > 2 (3Z- 4*)/48£7, £(£Z) = fP/192EI.
(b) Both ends supported:
£ = fz(3P - 4# 2 )/48£/, £(|Z) = //3/48£7.
The rod is symmetrical about its mid-point, so that the functions £(#) in \l ^ z s$ / are
obtained simply by replacing z by l—z.
Problem 3. The same as Problem 2, but for a rod clamped at one end (z = 0) and free
at the other end (z = I), to which a force/ is applied.
Solution. At all points of the rod F = constant = /, so that £'" == —f/EI. Using the
conditions £ = 0, £' = for z = 0, £" = for z = I, we obtain
£ = fz*(3l-z)l6EI, £(/) = /73/3E7.
Problem 4. Determine the shape of a rod with fixed ends, bent by a couple at its
mid-point.
Solution. At all points of the rod £ (lv) = 0, and at z = \l the moment M = Eli" has
a discontinuity equal to the moment m of the applied couple. The results are:
(a) Both ends clamped:
£ = mz\l -2z)l8EIl for ^ z ^ \l,
£ = -m(l-zf[l-2(l-z)]ISEIl for #<*</.
(b) Both ends hinged:
£ = mz(l 2 -4z 2 )j24EIl for < z ^ £/,
£ = -m(/-#)[Z 2 -4(Z-#) 2 ]/24£/Z for J/ < z < Z.
§20 Small deflections of rods 95
The rod is bent in opposite directions on the two sides of z = \l.
Problem 5. The same as Problem 4, but for the case where one end is clamped and the
other end free, the couple being applied at the latter end.
Solution. At all points of the rod M = Elt," = m, and at z = we have £ = 0, £' = 0.
The shape is given by £ = mz 2 l2EI,
Problem 6. Determine the shape of a circular rod with hinged ends stretched by a force
T and bent by a force / applied at its mid-point.
Solution. On the segment < z =£ J/ the shearing force is £/, so that (20.15) gives the
equation
£"-T£'IEI = -fj2EI.
The boundary conditions are £ = C = for z = and I; £' = for z = \l (since £' is
continuous). The shape of the rod (in the segment s$ z ^ \V) is given by
c = n z _ sinh ** ), k = v(w.
fe 2r\ kcoshiki!' vv
For small ft this gives the result obtained in Problem 2 (b). For large k it becomes £ = fzjlT,
i.e., in accordance with equations (20.17), a flexible wire under a force / takes the form of
two straight pieces intersecting at z = \l.
If the force T is due to the stretching of the rod by the transverse force, it must be deter-
mined by formula (20.16). Substituting the above result, we obtain the equation
1 r3 1 1 3 1 1 8E 2 / 3
— l-+-tanh 2 -&/ tanh *
H-
*»L2 2 2 kl 2 J PS '
which determines T as an implicit function off.
Problem 7. A circular rod of infinite length lies in an elastic substance, i.e. when it is
bent a force K = — oc£ proportional to the deflection acts on it. Determine the shape of the
rod when a concentrated force /acts on it.
Solution. We take the origin at the point where the force / is applied. The equation
E/£(iv) _ _ a £ holds everywhere except at z = 0. The solution must satisfy the condition
J=0at2= ±oo, and at z = £' and £" must be continuous; the difference between the
shearing forces F = —Eli"' for * -»> 0+ and z ->- 0— must be/. The required solution is
Problem 8. Derive the equation of equilibrium for a slightly bent thin circular rod which,
in its undeformed state, is an arc of a circle and is bent in its plane by radial forces.
Solution. Taking the origin of polar co-ordinates r, <f> at the centre of the circle, we write
the equation of the deformed rod as r = a + £(<£), where a is the radius of the arc and I a small
radial displacement. Using the expression for the radius of curvature in polar co-ordinates,
we find as far as the first order in £
1 r 2_ rr " + 2r'2 1 £+£"
R ( r 2 + r '2)3/2 a a 2
where the prime denotes differentiation with respect to <f>. According to (18.11), the elastic
bending energy is
r 1 1 1\ 2 EI r
96
The Equilibrium of Rods and Plates
§20
^o being the angle subtended by the arc at its centre. The equation of equilibrium is obtained
from the variational principle
00
SF r od- jSCKradfi** 0,
where K r is the external radial force per unit length, with the auxiliary condition
00
JCd^-0,
which is, in this approximation, the statement of the fact that the total length of the rod is
unchanged, i.e. it undergoes no general extension. Using Lagrange's method, we put
4>o
&Frod- jaKM d<f> + aoLJ8£ d<f> = 0,
where a is a constant. Varying the integrand in F to< i and integrating the 8 £" term twice by
parts, we obtain
f{— (£ + 2^ + £ (iv) )-«^r + aaW d<f> +
+^[«+D8n-^[«'+r)8a = o.
a 3 a 3
Hence we find the equation of equilibriumt
EI(P*>+2C" + Qla*-Kr+* = 0, (1)
the shearing force F= —EI(t,'+t,'")la*, and the bending moment M = £!/({ + £")/«*;
cf. the end of §20. The constant a is determined from the condition that the rod as a whole
is not stretched.
Problem 9. Determine the deformation of a circular ring bent by two forces / applied
along a diameter (Fig. 18).
t In the absence of external forces, K r = and a = 0; the non-zero solutions of the resulting
homogeneous equation correspond to a simple rotation or translation of the whole rod.
§21 The stability of elastic systems 97
Solution. Integrating equation (1), Problem 8, along the circumference of the ring, we
have 27T<xa = J K r a d<f> = 2/. We have equation (1) with K, = everywhere except at
<f> — and <f> — tt:
£Uv) + 21" + 1 +fa?J7TEI = 0.
The required deformation of the ring is symmetrical about the diameters AB and CD, and
so we must have £' = at A, B, C and D. The difference in the shearing forces for <f> -> ±
must be/. The solution of the equation of equilibrium which satisfies these conditions is
< IT.
£ = ( — h-<£cos<£ — 7rcos</> — sin<£), <: <f>
In particular, the points .4 and B approach through a distance
IROHJWI-y;-;)-
§21. The stability of elastic systems
The behaviour of a rod subject to longitudinal compressing forces is the
simplest example of the important phenomenon of elastic instability, first
discovered by L. Euler.
In the absence of transverse bending forces K x , K y , the equations of
equilibrium (20.14) for a compressed rod have the evident solution
X = Y = 0, which corresponds to the rod's remaining straight under a
longitudinal force |T|. This solution, however, gives a stable equilibrium
of the rod only if the compressing force | T\ is less than a certain critical value
Tor- For | T\ < T CT , the straight rod is stable with respect to any small pertur-
bation. In other words, if the rod is slightly bent by some small force, it will
tend to return to its original position when that force ceases to act.
If, on the other hand, \T\ > T CT , the straight rod is in unstable equilibrium.
An infinitesimal bending suffices to destroy the equilibrium, and a large
bending of the rod results. It is clear that, if this is so, the compressed rod
cannot actually remain straight.
The behaviour of the rod after it ceases to be stable must satisfy the equa-
tions for bending with large deflections. The value T cr of the critical load,
however, can be obtained from the equations for small deflections. For
\T\ = T cr , the straight rod is in neutral equilibrium. This means that, besides
the solution X = Y = 0, there must also be states where the rod is slightly
bent but still in equilibrium. Hence the critical value of T CI is the value of
|T| for which the equations
EI 2 X(W+ \T\X" = 0, EhY<™+ \T\Y" = (21.1)
have a non-zero solution. This solution gives also the nature of the deforma-
tion of the rod immediately after it ceases to be stable.
98 The Equilibrium of Rods and Plates §21
The following Problems give some typical cases of the loss of stability in
various elastic systems.
PROBLEMS
Problem 1 . Determine the critical compression force for a rod with hinged ends.
Solution. Since we are seeking the smallest value of \T\ for which equations (21.1) have
a non-zero solution, it is sufficient to consider only the equation which contains the smaller
of /j and I 2 . Let I % < I x . Then we seek a solution of the equation EI z X (lv ^-\-\T\X" =
in the form X = A+Bz+C sin kz+D cos kz, where k = \/(\T\/EI 2 ). The non-zero
solution which satisfies the conditions X = X" = for z = and z = / is X = C sin kz,
with sin kl = 0. Hence we find the required critical force to be T cr = tt'EIJP. On ceasing
to be stable, the rod takes the form shown in Fig. 19a.
(a)
Fig. 19
Problem 2. The same as Problem 1, but for a rod with clamped ends (Fig. 19b).
Solution. T cr = WEIJl 2 .
Problem 3. The same as Problem 1, but for a rod with one end clamped and the other
free (Fig. 19c).
Solution. T cr = n*EJJW.
Problem 4. Determine the critical compression force for a circular rod with hinged ends
in an elastic medium (see §20, Problem 7).
Solution. The equations (21.1) must now be replaced by EIXW + \T\X"+ctX = 0.
A similar treatment gives the solution X = A sin nnzjl,
tPEI/ od* \
Z2 \ nhflEl)
where n is the integer for which T cr is least. When a is large, n > 1, i.e. the rod exhibits
several undulations as soon as it ceases to be stable.
Problem 5. A circular rod is subjected to torsion, its ends being clamped. Determine
the critical torsion beyond which the straight rod becomes unstable.
Solution. The critical value of the torsion angle is determined by the appearance of
non-zero solutions of the equations for slight bending of a twisted rod. To derive these
equations, we substitute the expression (19.7) M = EltXdt/dl+Crt, where t is the constant
torsion angle, in equation (19.3). This gives
d 2 t dt
§21 The stability of elastic systems 99
We differentiate ; since the bending is not large, t may be regarded as a constant vector t
along the axis of the rod (the #-axis) in differentiating the first and third terms. Since also
dF/d/ = (there being no external forces except at the ends of the rod), we obtain
d3t d 2 t
or, in components,
Y^-kX'" = 0,
X^+kY'" = 0,
where k = CrjEI. Taking as the unknown function £ = X+iY,we obtain £< lv )— ii<£'" = 0.
We seek a solution which satisfies the conditions £ = 0, £' = for z = and z = I, in the
form £ = a(l+iKZ— e iKZ )+bz*, and obtain as the compatibility condition of the equations
for a and b the relation e iKl = (2+iW)/(2— tW), whence \kI — tan \kI. The smallest root
of this equation is \kI = 4-49, so that r cr = S-9SEIJCI.
Problem 6. The same as Problem 5, but for a rod with hinged ends.
Solution. In this case we have £ = a(l— e iKZ — i^z^+bz, where k is given by
e iKl = 1, i.e. kI = 2tt.
Hence the required critical torsion angle is t ci = 2-itEIJCI.
Problem 7. Determine the limit of stability of a vertical rod under its own weight, the
ower end being clamped.
Solution. If the longitudinal stress F s = T varies along the rod, dF t /dl ^ in the
first term of (20.1), and equations (20.14) are replaced by
hEX^-{TX')'-K x = 0,
hEYW-^TYJ-Ky = 0.
In the case considered, there are no transverse bending forces anywhere in the rod, and
T = — q(l— z), where q is the weight of the rod per unit length and z is measured from the
lower end. Assuming that I t < I lt we consider the equation
hEX'" = TX' = -q{l-z)X'\
illy. The general integral of 1
for z = I, X'" — automatically. The general integral of this equation for the function
« = X' is
where
The boundary conditions X' = for z — and X" = for z — I give for the function
«(i?) the conditions u = for t) = 170 = $V(ql 3 JEh)> u'r] 113 = for 17 = 0. In order to satisfy
these conditions we must put b = and J-i(-rjo) = 0- The smallest root of this equation
is 170 = 1 -87, and so the critical length is l cr ~ 1 -98(£/ 2 /g) 1/3 .
Problem 8. A rod has an elongated cross-section, so that I t ^> I x . One end is clamped
and a force/ is applied to the other end, which is free, so as to bend it in the principal *«-plane
(in which the flexural rigidity is EI 2 ). Determine the critical force / cr at which the rod bent
in a plane becomes unstable and the rod is bent sideways (in the yz-plane), at the same time
undergoing torsion.
100 The Equilibrium of Rods and Plates §21
Solution. Since the rigidity EI % is large compared with EI X (and with the torsional
rigidity C),t the instability as regards sideways bending occurs while the deflection in the
war-plane is still small. To determine the point where instability sets in, we must form the
equations for slight sideways bending of the rod, retaining the terms proportional to the
products of the force / in the xsr-plane and the small displacements. Since there is a concen-
trated force only at the free end of the rod, we have F = f at all points, and at the free end
(ar = I) the moment M = 0; from formula (19.6) we find the components of the moment
relative to a fixed system of co-ordinates x, y, z: M x = 0, M v = {l—z)f, M z — (Y— Y )f,
where Y — Y(l). Taking the components along co-ordinate axes £ , t], £ fixed at each point
to the rod, we obtain as far as the first-order terms in the displacements Mg — <f>(l—z)f,
M n = (/— z)f, M c = (l-z)fd Y/dz+f(Y— Y„), where <f> is the total angle of rotation of a
cross-section of the rod under torsion; the torsion angle t = d<f>ldz is not constant along
the rod. According to (18.6) and (18.9), however, we have for a small deflection
M g = -EhY", M v = EI 2 X", M^ = Cf ;
comparing, we obtain the equations of equilibrium
EhX" = {l-z)f y EhY" = -<f>(l-z)f,
Ccf>' = (l-z)fY' + (Y-Y )f.
The first of these equations gives the main bending of the rod, in the xz-plane ; we require
the value of/ for which non-zero solutions of the second and third equations appear. Eliminat-
ing Y, we find
<f>" + k\l- zf $ = 0, £2 = pfEhC.
The general integral of this equation is
* = «V(*- *!/*&*('- *)»]+ V('-*)/-i[P('-*) 2 ].
At the clamped end (z = 0) we must have <f> = 0, and at the free end the twisting moment
C<f>' = 0. From the second condition we have a = 0, and then the first gives J-^kl 2 ) = 0.
The smallest root of this equation is %kl 2 = 2-006, whence f CT = 4 -01 -\/^EIxC) 1 7*.
t For example, for a narrow rectangular cross-section of sides b and h (i > h), we have
EI X = bh 3 Ell2, ET t = VhE\\2, C = bh 3 fil3.
CHAPTER III
ELASTIC WAVES
§22. Elastic waves in an isotropic medium
If motion occurs in a deformed body, its temperature is not in general
constant, but varies in both time and space. This considerably complicates
the exact equations of motion in the general case of arbitrary motions.
Usually, however, matters are simplified in that the transfer of heat from
one part of the body to another (by simple thermal conduction) occurs very
slowly. If the heat exchange during times of the order of the period of
oscillatory motions in the body is negligible, we can regard any part of the
body as thermally insulated, i.e. the motion is adiabatic. In adiabatic defor-
mations, however, ow is given in terms of w$& by the usual formulae, the
only difference being that the ordinary (isothermal) values of E and a must be
replaced by their adiabatic values (see §6). We shall assume in what follows
that this condition is fulfilled, and accordingly E and a in this chapter will be
understood to have their adiabatic values.
In order to obtain the equations of motion for an elastic medium, we must
equate the internal stress force daucjdxjc to the product of the acceleration
tit and the mass per unit volume of the body, i.e. its density p :
pu t = da ik fdx k . (22.1)
This is the general equation of motion.
In particular, the equations of motion for an isotropic elastic medium can
be written down at once by analogy with the equation of equilibrium (7.2).
We have
F E
pu = A u + grad div u. (22.2)
2(1 + a) 2(l + <r)(l-2a) S V
Since all deformations are supposed small, the motions considered in the
theory of elasticity are small elastic oscillations or elastic waves. We shall
begin by discussing a plane elastic wave in an infinite isotropic medium, i.e.
a wave in which the deformation u is a function only of one co-ordinate
{x, say) and of the time. All derivatives with respect to y and z in equations
(22.2) are then zero, and we obtain for the components of the vector u the
equations
dhi x 1 d 2 u x ^ d 2 u y 1 d 2 u y ,„„ „v
— - = 0, — = (22.3)
dx 2 ci 2 dt 2 dx 2 Ct 2 dt 2
101
102 Elastic Waves §22
(the equation for u z is the same as that for u y ) ; heref
Cl ~ Vp(l + a)(l-2a)' Ct = fe' (22 ' 4)
Equations (22.3) are ordinary wave equations in one dimension, and the
quantities ci and c t which appear in them are the velocities of propagation of
the wave. We see that the velocity of propagation for the component u x is
different from that for u y and u z .
Thus an elastic wave is essentially two waves propagated independently.
In one (u x ) the displacement is in the direction of propagation; this is called
the longitudinal wave, and is propagated with velocity cj. In the other wave
(%> «z) the displacement is in a plane perpendicular to the direction of propa-
gation; this is called the transverse wave, and is propagated with velocity ct.
It is seen from (22.4) that the velocity of longitudinal waves is always greater
than that of transverse waves: we always have J
c t > V(4/3)^. (22.5)
The velocities ci and ct are often called the longitudinal and transverse veloci-
ties of sound.
We know that the volume change in a deformation is given by the sum of
the diagonal terms in the strain tensor, i.e. by ua = div u. In the transverse
wave there is no component u x , and, since the other components do not
depend on y or z, div u = for such a wave. Thus transverse waves do not
involve any change in volume of the parts of the body. For longitudinal
waves, however, div u ^ 0, and these waves involve compressions and
expansions in the body.
The separation of the wave into two parts propagated independently with
different velocities can also be effected in the general case of an arbitrary
(not plane) elastic wave in an infinite medium. We rewrite equation (22.2) in
terms of the velocities ci and ct :
u = c t 2 A u + (q 2 - c t 2 ) grad div u. (22.6)
We then represent the vector u as the sum of two parts :
u = ui + ut, (22.7)
of which one satisfies
div u t = (22.8)
and the other satisfies
curl ui = 0. (22.9)
We know from vector analysis that this representation (i.e. the expression of
t We may give also expressions for Cj and ct in terms of the moduli of compression and rigidity
and the Lam6 coefficients: c, = V{(3i^+V)/3p} = V{ (A+ 2fi)/p }, c t = V(m/p)-
J Since a actually varies only between and J (see the second footnote to §5), we always have
§22 Elastic waves in an isotropic medium 103
a vector as the sum of the curl of a vector and the gradient of a scalar) is
always possible.
Substituting u = uj + u f in (22.6), we obtain
Uj+ii* = c t 2 A(ui + u t ) + (ci 2 -ct 2 ) grad divuj. (22.10)
We take the divergence of both sides. Since div u t = 0, the result is
diviij = c t 2 A div ui + (c^-c t 2 ) A divuj,
or div(ui-ci 2 /\ui) = 0. The curl of the expression in parentheses is also
zero, by (22.9). If the curl and divergence of a vector both vanish in all
space, that vector must be zero identically. Thus
— - 'Muz = 0. (22.11)
dp
Similarly, taking the curl of equation (22.10) we have, since the curls of u t
and of any gradient are zero, curl (u t -c t 2 A^t) = 0. Since the divergence
of the expression in parentheses is also zero, we obtain an equation of the
same form as (22.11):
— - tfAut = 0. (22.12)
dt 2
Equations (22.11) and (22.12) are ordinary wave equations in three dimen-
sions. Each of them represents the propagation of an elastic wave, with
velocity c\ and c t respectively. One wave (u t ) does not involve a change in
volume (since div u t = 0), while the other (u{) is accompanied by volume
compressions and expansions.
In a monochromatic elastic wave, the displacement vector is
u = re{uo(r)*-^}, (22.13)
where uo is a function of the co-ordinates which satisfies the equation
c t 2 A uo + {ci 2 - c t 2 ) grad div uo + <u 2 uo = 0, (22. 14)
obtained by substituting (22.13) in (22.6). The longitudinal and transverse
parts of a monochromatic wave satisfy the equations
Aui + J^ui = 0, Au, + £t 2 ut= 0, (22.15)
where ki = co/cj, k t = cojc t are the wave numbers of the longitudinal and
transverse waves.
Finally, let us consider the reflection and refraction of a plane mono-
chromatic elastic wave at the boundary between two different elastic media.
It must be borne in mind that the nature of the wave is in general changed
when it is reflected or refracted. If a purely transverse or purely longitudinal
wave is incident on a surface of separation, the result is a mixed wave con-
taining both transverse and longitudinal parts. The nature of the wave
104
Elastic Waves
§22
remains unchanged (as we see from symmetry) only when it is incident
normally on the surface of separation, or when a transverse wave whose
oscillations are parallel to the plane of separation is incident (at any angle).
The relations giving the directions of the reflected and refracted waves can
be obtained immediately from the constancy of the frequency and of the
tangential components of the wave vector, f Let 6 and 6' be the angles of
incidence and reflection (or refraction) and c, c' the velocities of the two waves.
Then
sin# c
~^¥ = V (22 ' 16)
sina c
For example, let the incident wave be transverse. Then c = ct\ is the
velocity of transverse waves in medium 1. For the transverse reflected wave
we have c' = c t \ also, so that (22.16) gives 6 = 0', i.e. the angle of incidence
is equal to the angle of reflection. For the longitudinal reflected wave,
however, c' = en, and so
sin 6 ca
sin 6' en
For the transverse part of the refracted wave c' = ct2> and for a transverse
incident wave
sin 6 en
sin d' c t 2
Similarly, for the longitudinal refracted wave
sin 9 ct\
sin 6' C12
PROBLEMS
Problem 1. Determine the reflection coefficient for a longitudinal monochromatic wave
incident at any angle on the surface of a body (with a vacuum outside).*
Fig. 20
Solution. When the wave is reflected, there are in general both longitudinal and trans-
verse reflected waves. It is clear from symmetry that the displacement vector in the trans-
verse reflected wave lies in the plane of incidence (Fig. 20, where n , nj and nj are unit
f See Fluid Mechanics, §65. The arguments given there are applicable in their entirety.
I The more general case of the reflection of sound waves from a solid-liquid interface, and the
similar problem of the reflection of a wave incident from a liquid on to a solid, are discussed by L. M.
Brekhovskikh, Waves in Layered Media, §4, Academic Press, New York 1960.
§22 Elastic waves in an isotropic medium 105
vectors in the direction of propagation of the incident, longitudinal reflected and transverse
reflected waves, and u , u t , Ut the corresponding displacement vectors). The total displace-
ment in the body is given by the sum (omitting the common factor e~ im for brevity)
u = A noe ik *- r +Ainie ik rr+A t aXnte ik f r ,
where a is a unit vector perpendicular to the plane of incidence. The magnitudes of the wave
vectors are k = k t = w/cj, kt = ojct, and the angles of incidence 9 and of reflection 9 U
&t are related by 0j = 9 , sin dt = (ct/cj) sin O . For the components of the strain tensor at
the boundary we obtain
u xx = ik<s{AQ + Ai) cos 2 #o + iAtkt cos 0* sin 0*, uu = iko(Ao + At),
u X y = iko(Ao — Ai) sin do cos 6o+ ^iAtk t (cos 2 6 t — sin 2 0*),
again omitting the common exponential factor. The components of the stress tensor can be
calculated from the general formula (5.11), which can here be conveniently written
o ilc = 2pC t 2 Uik + p(cp -2c t 2 )uu8 ik .
The boundary conditions at the free surface of the medium are a<k«* = 0, whence
<*xx — a yx — 0,
giving two equations which express Ai and At in terms of A . The result is
c t 2 sin 20* sin 20 o - c t 2 cos 2 20*
A x = A
A t = -A G
c t 2 sin 20* sin 20 o + Q 2 cos 2 2d t
2ciCt sin 20o cos 20*
c t 2 sin 20* sin 20 o + c t 2 cos 2 20*
For 9 = we have A t = — A , At = 0, i.e. the wave is reflected as a purely longitudinal
wave. The ratio of the energy flux density components normal to the surface in the reflected
and incident longitudinal waves is Ri = |^4jM | a - The corresponding ratio for the reflected
transverse wave is
Rt =
c t cos 0*
A
|2
Ci COS 00
The sum of R t and Rt is, of course, 1.
Problem 2. The same as Problem 1, but for a transverse incident wave (with the oscilla-
tions in the plane of incidence). -f
Solution. The wave is reflected as a transverse and a longitudinal wave, with 0< = o ,
Ct sin 9 % = Cj sin 9 . The total displacement vector is
u = aXno^oe* k »- r +n^ze <k '' r +aXn*^*e* k '-«\
The expressions for the amplitudes of the reflected waves are
A t c t 2 sin 20; sin 20 o - ci 2 cos 2 20 o
Aq ~~ c t 2 sin 20j sin 20 o + ci 2 cos 2 20 o '
Ai 2ciCf sin 20o cos 20o
Aq ~ c t 2 sin 20; sin 20 o + ci 2 cos 2 20 o '
t If the oscillations are perpendicular to the plane of incidence, the wave is entirely reflected a* a
wave of the same kind, and so Rt = 1 .
106 Elastic Waves §23
Problem 3. Determine the characteristic frequencies of radial vibrations of an elastic
phere of radius R.
Solution. We take spherical polar co-ordinates, with the origin at the centre of the sphere.
For radial vibrations, u is along the radius, and is a function of r and t only. Hence curl u = 0.
We define the displacement "potential" 4> by u r = u = 8<f>ldr. The equation of motion,
expressed in terms of <f>, is just the wave equation ci s A<A = $, or, for oscillations periodic in
time (~e~ i<Jit ),
The solution which is finite at the origin is <j> = (A/r) sin kr (the time factor is omitted). The
radial stress is
a rr = pi (ci 2 — 2ct 2 )uu + 2c t 2 U rr
= p\{c l 2 -2c t 2 )/\<l> + 2c t 2 <l>"
or, using (1),
Orrlp= -o> 2 <f>-W<f>'lr. (2)
The boundary condition a rr (R) = leads to the equation
tan kR 1
kR l-(kRcil2c t ) 2 ' (3)
whose roots determine the characteristic frequencies w = kci of the vibrations.
Problem 4. Determine the frequency of radial vibrations of a spherical cavity in an infinite
elastic medium for which ci !> c t (M. A. Isakovich 1949).
Solution. In an infinite medium, radial oscillations of the cavity are accompanied by the
emission of longitudinal sound waves, leading to loss of energy and hence to damping of the
oscillations. When ci ^> c t (i.e. K ^> /x), this emission is weak, and we can speak of the charac-
teristic frequencies of oscillations with a small coefficient of damping.
We seek a solution of equation (1), Problem 3, in the form of an outgoing spherical wave
ff> = Ae ikr lr, k = oi\ci and, using (2), obtain from the boundary condition o rr (R) = the
result {kRcijctY = 4(1 —ikR). Hence, when c? ^> c t ,
lc t I c t \
co — — 1 1— i — .
R\ ci)
The real part of ou gives the characteristic frequency of oscillation ; the imaginary part gives
the damping coefficient. In an incompressible medium (ci ~> oo) there would of course be no
damping. These vibrations are specifically due to the shear resistance of the medium (fi #0).
Il should be noticed that they have kR = 2ct\ci <^. 1 , i.e. the corresponding wavelength is
large compared with R ; it is interesting to compare this with the result for vibrations of an
elastic sphere, where with ci ^> ct the first characteristic frequency is given by (3) : kR = -it.
§23. Elastic waves in crystals
The propagation of elastic waves in anisotropic media, i.e. in crystals, is
more complicated than for the case of isotropic media. To investigate such
waves, we must return to the general equations of motion piii = dotje/dxk
and use for o-^ the general expression (10.3) 0% = \MmUim- According to
§23 Elastic waves in crystals 107
what was said at the beginning of §22, A^zm always denotes the adiabatic
moduli of elasticity.
Substituting for o^ in the equations of motion, we obtain
duim v d l dui du
piii = XiMm— — = l^iklm
OXjc OXfc
l dui du m \
d 2 u t d 2 u m
= iMklmrz — H f Mklmr
dx]cdx m dxjcdxi
Since the tensor Xikim is symmetrical with respect to the suffixes / and m
we can interchange these in the first term, which then becomes identical
with the second term. Thus the equations of motion are
d 2 Um
piii = Xikim— — - — • (23.1)
OXkOXi
Let us consider a monochromatic elastic wave in a crystal. We can seek
a solution of the equations of motion in the form Ui = uoie i{k ' r ~ a)t \ where the
uoi are constants, the relation between the wave vector k and the frequency (o
being such that this function actually satisfies equation (23.1). Differentiation
of m with respect to time results in multiplication by — ico, and differentia-
tion with respect to xjc leads to multiplication by ikjc. Hence the above substi-
tution converts equation (23.1) into pcohii = AijcimkkkiUm. Puttings = 8i m Um,
we can write this as
(p(o 2 him-\ucimkkki)U m = 0. (23.2)
This is a set of three homogeneous equations of the first degree for the
unknowns u%, u y , u z . Such equations have non-zero solutions only if the
determinant of the coefficients is zero. Thus we must have
\hkimkkh— poi 2 8 im \ — 0. (23.3)
This is a cubic equation in co 2 . It has three roots, which are in general
different. Each root gives the frequency as a function of the wave vector k.f
Substituting each in turn in equation (23.2), we obtain equations giving
the components of the corresponding displacement w* (since the equations
are homogeneous, of course, only the ratios of the three components «* are
obtained, and not their absolute values, so that all the m can be multiplied
by an arbitrary constant).
The velocity of propagation of the wave (the group velocity) is given by the
derivative of the frequency with respect to the wave vector. In an isotropic
body, the frequency is proportional to the magnitude of k, and so the direc-
tion of the velocity U = dco/dk is the same as that of k. In crystals this
relation does not hold, and the direction of propagation of the wave is there-
fore not the same as that of its wave vector.
t In an isotropic body, equation (23.3) gives the result previously obtained: one root u>* = c*k*
and two coincident roots to 2 = C( 2 A a .
108 Elastic Waves §23
It is seen from equation (23.3) that o> is a homogeneous function, of degree
one, of the components ki ; if the unknown quantity is taken as the ratio cufk,
the coefficients in the equation do not depend on k. Hence the velocity of
propagation 9cu/8k is a homogeneous function, of degree zero, of ki. Thus
the velocity of a wave is a function of its direction, but not of its frequency.
Since there are three possible relations between o» and k for any direction
in the crystal, there are in general three different velocities of propagation
of elastic waves. These velocities are the same only in a few exceptional
directions.
In an isotropic medium, purely longitudinal and purely transverse waves
correspond to two different velocities of propagation. In a crystal, on the
other hand, to each velocity of propagation there corresponds a wave in which
the displacement vector has components both parallel and perpendicular
to the direction of propagation.
Finally, we may notice the following. For any given wave vector k in a
crystal there can be three waves, with different frequencies and velocities of
propagation. It is easy to see that the displacement vectors u in these three
waves are mutually perpendicular. For, when k is given, equation (23.3) may
be regarded as determining the principal values pco 2 of a tensor of rank two,
Xiicimkkki, which is symmetrical with respect to the suffixes i, m.f Equations
(23.2) then give the principal axes of this tensor, which we know are mutually
perpendicular.
PROBLEM
Determine the frequency as a function of the wave vector for elastic waves propagated in a
crystal of the hexagonal system.
Solution. The non-zero components of the tensor A»&jm in the co-ordinates x, y, z are
related to those in the co-ordinates £, ■q, z (see §10) by
Axxxx — Ayyyy = a + b, Axxyy — a ~ b y Axyxy = b,
^XXZZ = Ayyzz = C, &XZXZ = AyZyZ = U, &ZZZZ = />
where we have put
The ar-axis is along the sixth-order axis of symmetry; the directions of the x and y axes are
arbitrary. We take the xz-plane such that it contains the wave vector k. Then k x = k sin 6,
k y — 0, kz = k cos 0, where 6 is the angle between k and the ar-axis. Forming the equation
(23.3) and solving it, we obtain three different dependences of to on k:
co! 2 = k 2 (bs'm2d + dcos 2 e)! P ,
k?
C02 3 2 = — {(a + b) sinW+f cosW+d± ^([(a+ b - d) sinW + (d-f) cosWf +
2 P
+ 4(c+rf) 2 sm 2 0cos 2 0)}.
t By the symmetry of the tensor \ncim, we have Xikimkkki = hktmikjcki = Xmihikicki- The latter
expression differs from Xmkiikjcki only by the naming of the suffixes k and /, so that the tensor
Xikimkkki has the symmetry stated.
§24 Surface waves 109
§24. Surface waves
A particular kind of elastic waves are those propagated near the surface
of a body without penetrating into it (Rayleigh waves). We write the equation
of motion in the form (22.11) and (22.12):
— - c*Au = 0, (24.1)
dt 2
where u is any component of the vectors uj, u t , and c is the corresponding
velocity c\ or c t , and srek solutions corresponding to these surface waves.
The surface of the elastic medium is supposed plane and of infinite extent.
We take this plane as the ry-plane; let the medium be in z < 0.
Let us consider a plane monochromatic surface wave propagated along the
#-axis. Accordingly u = «<(**-**>/(*). Substituting this expression in (24.1), we
obtain for the function f(z) the equation
dy
d*2
If k 2 - ofijc 2 < 0, this equation gives a periodic function /, i.e. we obtain
an ordinary plane wave which is not damped inside the body. We must
therefore suppose that k 2 -w 2 ]c 2 > 0. Then the solutions for /are
f(z) = constant x exp { ± J \k 2 -m.
The solution with the minus sign would correspond to an unlimited increase
in the deformation for z -» — oo. This solution is clearly impossible, and
so the plus sign must be taken.
Thus we have the following solution of the equations of motion:
u = constant x «***-**>««, (24.2)
where
K = ^(k 2 -a> 2 !c 2 ). (24.3)
It corresponds to a wave which is exponentially damped towards the interior
of the medium, i.e. is propagated only near the surface. The quantity k
determines the rapidity of the damping.
The true displacement vector u in the wave is the sum of the vectors uj and
Uf, the components of each of which satisfy the equation (24.1) with c = ci
for ui and ct for u$. For volume waves in an infinite medium, the two parts
are independently propagated waves. For surface waves, however, this
division into two independent parts is not possible, on account of the boundary
conditions. The displacement vector u must be a definite linear combination
of the vectors u* and Uf . It should also be mentioned that these latter vectors
have no longer the simple significance of the displacement components
parallel and perpendicular to the direction of propagation.
110 Elastic Waves §24
To determine the linear combination of the vectors uj and ut which gives
the true displacement u, we must use the conditions at the boundary of the
body. These give a relation between the wave vector k and the frequency &>,
and therefore the velocity of propagation of the wave. At the free surface
we must have cr^n fc = 0. Since the normal vector n is parallel to the s-axis,
it follows that a xz = o yz = a zz = 0, whence
U X z = 0, Uyz = 0, v(u Z x + Uyy) + (l-v)Uzz = 0. (24.4)
Since all quantities are independent of the co-ordinate y, the second of
these conditions gives
1 iduy du z \
Uyz = - \-^- + -J~) = % du vl dz = °*
Using (24.2), we therefore have
u y = 0. (24.5)
Thus the displacement vector u in a surface wave is in a plane through the
direction of propagation perpendicular to the surface.
The transverse part u$ of the wave must satisfy the condition (22.8)
div M t = 0, or
du tx du tz
.+-—. = 0.
dx dz
The dependence of utx and Utz on x and z is determined by the factor
e ikx+K t z , where t<t is given by the expression (24.3) with c = ct, i.e.
K = \/(k 2 —C0 2 jCt 2 ).
Hence the above condition leads to the equation
ikutx+KtUtz = 0, or ut x /ut z = —K t Jik.
Thus we can write
u tx = K t ae ikx + K t z - i(0 t y u tz = -ikae ikx+K t z - icot , (24.6)
where a is some constant.
The longitudinal part u* satisfies the condition (22.9) curl uj = 0, or
dui x dui z ^
dz dx
whence
ikui z —KiUix = (ki = \Z[k 2 —co 2 Jci 2 ]).
Thus we must have
uix = kbe ikx+K i z - ia,t , ui z = -iKibe ikx+K i z - iu)t , (24.7)
where b is a constant.
§24 Surface waves 111
We now use the first and third conditions (24.4). Expressing um in terms
of the derivatives of U{, and using the velocities ci, ct, we can write these
conditions as
du x du z
+ = 0,
dZ dX (24.8)
du z du x
oz ox
Here we must substitute u x = ui x + ut%, u z = ttiz + utz- The result is that
the first condition (24.8) gives
a(k? + Kt*) + 2bkKi = 0. (24.9)
The second condition leads to the equation
2ac?K& + b[c?(ic?-k*) + 2c?k*] = 0.
Dividing this equation by Ct 2 and substituting
we can write it as
2aic£ + b(&+K?) = 0. (24.10)
The condition for the two homogeneous equations (24.9) and (24.10)
to be compatible is (k 2 + Kt 2 ) 2 = 4k 2 K t Ki or, squaring and substituting
the values of K t 2 and ki 2 ,
I to 2 \ 4 / co 2 \ / ft) 2 \
From this equation we obtain the relation between o> and k. It is convenient
to put
co = cM\ (24.12)
k 8 then cancels from both sides of the equation, and, expanding, we obtain
for £ the equation
$«-8£*+S€ 2 (3-2—\-l6(\ - —\ = 0. (24.13)
Hence we see that £ depends only on the ratio ctjci, which is a constant
characteristic of any given substance and in turn depends only on Poisson's
ratio :
CtjCi= V((l-2<r)/2(l-a)}.
112
Elastic Waves
§24
The quantity £ must, of course, be real and positive, and £ < 1 (so that
Kt and ki are real). Equation (24.13) has only one root satisfying these con-
ditions, and so a single value of f is obtained for any given value of ct/ci.
Thus, for both surface waves and volume waves, the frequency is pro-
portional to the wave number. The proportionality coefficient is the velocity
of propagation of the wave,
U = c&. (24.14)
This gives the velocity of propagation of surface waves in terms of the
velocities ct and ci of the transverse and longitudinal volume waves. The
ratio of the amplitudes of the transverse and longitudinal parts of the wave
is given in terms of £ by the formula
2-|2
2V(1-| 2 )
(24.15)
The ratio ct\ci actually varies from \jy/2 to for various substances,
corresponding to the variation of a from to \\ £ then varies from 0-874 to
0-955. Fig. 21 shows a graph of f as a function of a.
100
0-95
0-90
085
PROBLEM
A plane- parallel slab of thickness h (medium 1) lies on an elastic half-space (medium 2).
Determine the frequency as a function of the wave number for transverse waves in the slab
whose direction of oscillation is parallel to its boundaries.
Solution. We take the plane separating the slab from the half-space as the xy-plane,
the half-space being in z < and the slab in s£ z <; h. In the slab we have
Uzl = u zl = 0, u yl = f{z)eM*-<>>t\
and in medium 2 a damped wave :
Ux2 = U z2 = 0, % 2 = ^eV« i(M) , K2 = V(k 2 -0 2 ICt2 2 ).
For the function f(z) we have the equation
/"+ K?f =0, Kl = V(«> 2 fCtl 2 -k*)
§25 Vibration of rods and plates ' 113
(we shall see below that k^ > 0), whence f(z) = B sin k x z+C cos k x z. At the free surface
of the slab {z — h) we must have o ty — 0, i.e. duyjdz = 0. At the boundary between the
two media (z = 0) the conditions are u yl = u vt , frdtij/Jdz = fi^du^dz, ^ and /i, being the
moduli of rigidity for the two media. From these conditions we find three equations for
A, B, C, and the compatibility condition is tan k-Ji = iH K ilfh. K i- This equation gives <o
as an implicit function of k ; it has solutions only for real #c x and k 2 , and so Ct% > (o/k > en.
Hence we see that such waves can be propagated only if c*2 > c«i.
§25. Vibration of rods and plates
Waves propagated in thin rods and plates are fundamentally different
from those propagated in a medium infinite in all directions. Here we are
speaking of waves of length large compared with the thickness of the rod or
plate. If the wavelength is small compared with this thickness, the rod or
plate is effectively infinite in all directions as regards the propagation of the
wave, and we return to the results obtained for infinite media.
Waves in which the oscillations are parallel to the axis of the rod or the
plane of the plate must be distinguished from those in which they are per-
pendicular to it. We shall begin by studying longitudinal waves in rods.
A longitudinal deformation of the rod (uniform over any cross-section), with
no external force on the sides of the rod, is a simple extension or compression.
Thus longitudinal waves in a rod are simple extensions or compressions
propagated along its length. In a simple extension, however, only the com-
ponent a zz of the stress tensor (the ar-axis being along the rod) is different
from zero; it is related to the strain tensor by a zz = Eu zz = Edu z jdz
(see §5). Substituting this in the general equation of motion pu z = 9<t z a;/3^a;»
we find
^_Z_^ = . (25.1)
This is the equation of longitudinal vibrations in rods. We see that it is an
ordinary wave equation. The velocity of propagation of longitudinal waves
in rods is
V(E/p). (25.2)
Comparing this with the expression (22.4) for q, we see that it is less than
the velocity of propagation of longitudinal waves in an infinite medium.
Let us now consider longitudinal waves in thin plates. The equations of
motion for such vibrations can be written down at once by substituting
— phdhiz/dt 2 and —phdhiyjdt 2 for P x and P y in the equilibrium equations
(13.4):
p d 2 u z 1 d 2 u x 1 d 2 u x 1 d 2 u y
~E~W = l-a 2 1hP + 2(\ + a) Hy* + 2{\-o)lh&y'
p d 2 u y 1 d 2 u y 1 B 2 Uy 1 d 2 u z
•+t— : — _ „ +•
(25.3)
E dt 2 1 - a 2 dy 2 2(1 + a) Bx 2 2(1 - a) dxdy
114 Elastic Waves §25
We take the case of a plane wave propagated along the #-axis, i.e. a wave in
which the deformation depends only on the co-ordinate x, and not on y.
Then equations (25.3) are much simplified, becoming
&U % E d*U X = d*Uy E d*Uy =
d& pO—a 2 ) d*2 ' dfi 2p(l + a) dx* ' ^ ' '
We thus again obtain wave equations. The coefficients are different for u x
and u y . The velocity of propagation of a wave with oscillations parallel to the
direction of propagation (u x ) is
ViElrfl-6*)]. (25.5)
The velocity for a wave (%) with oscillations perpendicular to the direction
of propagation (but still in the plane of the plate) is equal to the velocity ct of
transverse waves in an infinite medium.
Thus we see that longitudinal waves in rods and plates are of the same
nature as in an infinite medium, only the velocity being different; as before,
it is independent of the frequency. Entirely different results are obtained for
bending waves in rods and plates, for which the oscillations are in a direction
perpendicular to the axis of the rod or the plane of the plate, i.e. involve
bending.
The equations for free oscillations of a plate can be written down at once
from the equilibrium equation (12.5). To do so, we must replace — P by the
acceleration £ multiplied by the mass ph per unit area of the plate. This
gives
9 2 £ Eh*
»w + W^) AH = °' (25 ' 6)
where A is the two-dimensional Laplacian.
Let us consider a monochromatic elastic wave, and accordingly seek a
solution of equation (25.6) in the form
£ = constant xe« k "- wi ), (25.7)
where the wave vector k has, of course, only two components, k x and k y .
Substituting in (25.6), we obtain the equation
-pco 2 + Eh*k*/12(l-o2) = 0.
Hence we have the following relation between the frequency and the wave
number :
a) = kW{Eh*ll2 P (l - <x2)}. (25.8)
Thus the frequency is proportional to the square of the wave number, whereas
in waves in an infinite medium it is proportional to the wave number itself.
§25 Vibration of rods and plates 115
Knowing the relation between the frequency and the wave number, we can
determine the velocity of propagation of the wave from the formula
U = dcojdk.
The derivatives of k 2 with respect to the components k Xi k y are respectively
2k x , 2k y . The velocity of propagation of the wave is therefore
U = kV{Eh*/3 P (l - a 2 )}. (25.9)
It is proportional to the wave vector, and not a constant as it is for waves in
a medium infinite in three dimensions.^
Similar results are obtained for bending waves in thin rods. The bending
deflections of the rod are supposed small. The equations of motion are
obtained by replacing — K x and — K y in the equations of equilibrium for a
slightly bent rod (20.4) by the product of the acceleration X or Y and the
mass pS per unit length of the rod (S being its cross-sectional area). Thus
P SX = EI y d*XJdz\ P SY = ElaPY/dz*. (25.10)
We again seek solutions of these equations in the form
X = constant x e«*z-<*« Y = constant x e i{kz - mt \
Substituting in (25.10), we obtain the following relations between the fre-
quency and the wave number :
co = kty(EIyl P S), co = kW(EIxlpS), (25.11)
for vibrations in the x and y directions respectively. The corresponding
velocities of propagation are
U<*> = Ik^iEIyfpS), W> = 2kV(EI x l P S). (25.12)
Finally, there is a particular case of vibration of rods called torsional
vibration. The corresponding equations of motion are derived by equating
Cdrjdz (see §18) to the time derivative of the angular momentum of the rod
per unit length. This angular momentum is pld<f>fdt y where d<j>jdt is the
angular velocity (<f> being the angle of rotation of the cross-section considered)
and / = J (x 2 +y 2 ) d/ is the moment of inertia of the cross-section about its
centre of mass; for pure torsional vibration each cross-section of the rod
performs rotary vibrations about its centre of mass, which remains at rest.
Putting t = d<f>/dz, we obtain the equation of motion in the form
Cd 2 <f>/dz 2 = pidmet 2 . (25.13)
f The wave number k = 2-ir/X, where A is the wavelength. Hence the velocity of propagation
should increase without limit as A tends to zero. This physically impossible result is obtained because
formula (25.9) is not valid for short waves.
116 Elastic Waves §25
Hence we see that the velocity of propagation of torsional oscillations along
the rod is
V(C/ P I). (25.14)
PROBLEMS
Problem 1. Determine the characteristic frequencies of longitudinal vibrations of a rod
of length /, with one end fixed and the other free.
Solution. At the fixed end (z = 0) we must have u t = 0, and at the free end (z = /)
<*« = Eutg = 0, i.e. duzldz = 0. We seek a solution of equation (25.1) in the form
# 2 = A cos(a>t + (x) sinks,
where k = <o\Z(p/E). From the condition at z = I we have cos kl — 0, whence the charac-
teristic frequencies are
eo = V(Elp)(2n+l)7rl2l,
n being any integer.
Problem 2. The same as Problem 1, but for a rod with both ends free or both fixed.
Solution. In either case <o = \/(E/p) tin/ 1.
Problem 3. Determine the characteristic frequencies of vibration of a string of length /.
Solution. The equation of motion of the string is
d*X oS d*X
L == 0;
cf. the equilibrium equation (20.17). The boundary conditions are that X — f or z —
and /. The characteristic frequencies are o> = ■y/(pS/T)nn/l.
Problem 4. Determine the characteristic transverse vibrations of a rod (of length /) with
clamped ends.
Solution. Equation (25.10), on substituting X = X (z) cos(a>t +<x), becomes
d^o/d* 4 = k*Xo,
where k 1 = uPpS/EIy. The general integral of this equation is
Xq = A cos kz + B sin kz+C cosh kz+D sinh kz.
The constants A, B, C and D are determined from the boundary conditions that X = dX/dz
= f or z = and /. The result is
Xq = Aftsmfcl— sinh «r/)(cos kz— cosh kz) —
— (cos kI— cosh «/)(sin kz — sinh kz)},
and the equation coskZ cosh#tZ = 1, the roots of which give the characteristic frequencies
The smallest characteristic frequency is
224 l Eh
COmin =
■ I Ely
V pS
/2 V P S
Problem 5. The same as Problem 4, but for a rod with supported ends.
§25 Vibration of rods and plates 117
Solution. In the same way as in Problem 4, we obtain X = A sin kz, and the frequencies
are given by sin*/ = 0, i.e. k = nn/l(n = 1, 2, ...). The smallest frequency is
y-87 I Ely
Problem 6. The same as Problem 4, but for a rod with one end clamped and the other
free.
Solution. We have for the displacement
Xo = A{(cos kI+ cosh k/)(cos kz— cosh kz)
+ (sin kI— sinh /c/)(sin kz — sinh kz)}
(the clamped end being at z — and the free end atz = I), and for the characteristic fre-
quencies the equation cos kI cosh kZ+1 = 0. The smallest frequency is
3-52 J Ely
COmin
3-5Z I Ely
= ~¥~J~p~S~'
P S
Problem 7. Determine the characteristic vibrations of a rectangular plate of sides a and b,
with its edges supported.
Solution. Equation (25.6), on substituting £ = l (x, y) cos(a>t+a), becomes
where #c* = 12p(l — a^wPjEh*. We take the co-ordinate axes along the sides of the plate.
The boundary conditions (12.11) become £ = 9 2 £/3* 8 = for x = and a,
for y = and b. The solution which satisfies these conditions is
Co = A sm(m7rx/a) s'mfoiry/b),
where m and n are integers. The frequencies are given by
E _ [m 2 w 2 l
12o(l-o- 2 ) 7
CO = h / 7T 2 — + — .
Vl2p(l-a2) [ a 2 #jj
Problem 8. Determine the characteristic frequencies for the vibration of a rectangular
membrane of sides a and b.
Solution. The equation for the vibration of a membrane is TA £ — phi', cf. the equili-
brium equation (14.9). The edges of the membrane must be fixed, so that £ — 0. The
corresponding solution for a rectangular membrane is
£ = A sm(niTTxla) sin(»7ry/#) cos cot,
where the characteristic frequencies are given by
Ttt 2 /m 2 n 2 \
ft) 2 = — — +— I,
ph W bV
m and n being integers.
118 Elastic Waves §26
Problem 9. Determine the velocity of propagation of torsional vibrations in a rod whose
cross-section is a circle, an ellipse, or an equilateral triangle, and in a rod in the form of a
long thin rectangular plate.
Solution. For a circular cross-section of radius R, the moment of inertia is / = \ttR*',
C is given in §16, Problem 1, and we find the velocity to be V(plp), which is the same as the
velocity ct.
Similarly (using the results of §16, Problems 2 to 4), we find for a rod of elliptical cross-
section the velocity [2a6/(a 2 +fc 2 )] vX^/p). for one with an equilateral triangular cross-section
V(3l*l5p), and for one which is a long rectangular plate (2h/d)\/(fjL/p). All these are less than a.
Problem 10. The surface of an incompressible fluid of infinite depth is covered by a thin
elastic plate. Determine the relation between the wave number and the frequency for waves
which are simultaneously propagated in the plate and near the surface of the fluid.
Solution. We take the plane of the plate as 2 = 0, and the x-axis in the direction of
propagation of the wave ; let the fluid be in z < 0. The equation of motion of the plate alone
would be
d% _ Em d*£
po Jfi ~ ~ 12(1-0*) a?
where p is the volume density of the plate. When the fluid is present, the right-hand side
of this equation must also include the force exerted by the fluid on unit area of the plate,
i.e. the pressure p of the fluid. The pressure in the wave, however, can be expressed in
terms of the velocity potential by p = —pd<f>ldt (we neglect gravity). Hence we obtain
a 2 £ _ Em wz, r d<f>i
poh ~dF = ~ 12(1 - CT 2) W " l Tt\z=o (1)
Next, the normal component of the fluid velocity at the surface must be equal to that of the
plate, whence
dydt = [ty/a*],..o. (2)
The potential <f> must satisfy everywhere in the fluid the equation
~+- J ~ = 0. (3)
dx 2 dz 2
We seek £ in the form of a travelling wave £ = £ e ikx ~ ia)t ; accordingly, we take as the
solution of equation (3) the surface wave <f> = <f> e i( - lcz ~ M '>e lcz , which is damped in the interior
of the fluid. Substituting these expressions in (1) and (2), we obtain two equations for ^
and £o» and the compatibility condition is
Eh^ &
12(1 -a 2 ) p + hpok
§26. Anharmonic vibrations
The whole of the theory of elastic vibrations given above is approximate
to the extent that any theory of elasticity is so which is based on Hooke's
law. It should be recalled that the theory begins from an expansion of the
elastic energy as a power series with respect to the strain tensor, which
includes terms up to and including the second order. The components of
§26 Anharmonic vibrations 119
the stress tensor are then linear functions of those of the strain tensor, and
the equations of motion are linear.
The most characteristic property of elastic waves in this approximation is
that any wave can be obtained by simple superposition (i.e. as a linear com-
bination) of separate monochromatic waves. Each of these is propagated
independently, and could exist by itself without involving any other motion.
We may say that the various monochromatic waves which are simultaneously
propagated in a single medium do not interact with one another.
These properties, however, no longer hold in subsequent approximations.
The effects which appear in these approximations, though small, may be of
importance as regards certain phenomena. They are usually called anharmonic
effects, since the corresponding equations of motion are non-linear and do
not admit simple periodic (harmonic) solutions.
We shall consider here anharmonic effects of the third order, arising from
terms in the elastic energy which are cubic in the strains. It would be too
cumbersome to write out the corresponding equations of motion in their
general form. However, the nature of the resulting effects can be ascertained
as follows. The cubic terms in the elastic energy give quadratic terms in the
stress tensor, and therefore in the equations of motion. Let us suppose that
all the linear terms in these equations are on the left-hand side, and all the
quadratic terms on the right-hand side. Solving these equations by the
method of successive approximations, we omit the quadratic terms in the
first approximation. This leaves the ordinary linear equations, whose solution
uo can be put in the form of a superposition of monochromatic travelling
waves: constant xe i{k ' r ~ (0t) , with definite relations between co and k. On
going to the second approximation, we must put u = uo + ui and retain only
the terms in uo on the right-hand sides of the equations (the quadratic terms).
Since uo, by definition, satisfies the homogeneous linear equations obtained
by putting the right-hand sides equal to zero, the terms in uo on the left-hand
sides will cancel. The result is a set of inhomogeneous linear equations for the
components of the vector ui, where the right-hand sides contain only known
functions of the co-ordinates and time. These functions, which are obtained
by substituting uo for u in the right-hand sides of the original equations, are
sums of terms each of which is proportional to
^•[(ki-k,).!— (o^-w,)*]
or
where coi, o>2, ki, k2 are the frequencies and wave vectors of any two mono-
chromatic waves in the first approximation.
A particular integral of linear equations of this type is a sum of terms
containing similar exponential factors to those in the free terms (the right-
hand sides) of the equations, with suitably chosen coefficients. Each such
term corresponds to a travelling wave with frequency coi ± a>2 and wave
120 Elastic Waves §26
vector ki + k2. Frequencies equal to the sum or difference of the frequencies
of the original waves are called combination frequencies.
Thus the anharmonic effects in the third order have the result that the set
of fundamental monochromatic waves (with frequencies a>i, a>2, ... and wave
vectors ki, k2, ...) has superposed on it other "waves" of small intensity,
whose frequencies are the combination frequencies such as to± + to2, and
whose wave vectors are such as ki±k2. We call these "waves" in quotation
marks because they are a correction effect and cannot exist alone except in
certain special cases (see below). The values to\ ± a>2 and ki + k2 do not in
general satisfy the relations which hold between the frequencies and wave
vectors for ordinary monochromatic waves.
It is clear, however, that there may happen to be particular values of a>i, ki
and a>2, k2 such that one of the relations for monochromatic waves in the
medium considered also holds for a>i+o>2 and ki + k2 (for definiteness, we
shall discuss sums and not differences). Putting toz = coi+102, ks = ki + k2,
we can say that, mathematically, 103 and k3 then correspond to waves which
satisfy the homogeneous linear equations of motion (with zero on the right-
hand side) in the first approximation. If the right-hand sides in the second
approximation contain terms proportional to £ < < k »' r-w » ( > > then a particular
integral will be a wave with the same frequency and an amplitude which
increases indefinitely with time.
Thus the superposition of two monochromatic waves with values of to±, ki
and co2, k2 whose sum C03, k3 satisfies the above condition leads, by the
anharmonic effects, to resonance: a new monochromatic wave (with para-
meters C03, h.3) is formed, whose amplitude increases with time and eventually
is no longer small. It is evident that, if a wave with 003, k3 is formed on super-
position of those with a>i, ki and C02, k2, then the superposition of waves with
wi, ki and 003, k3 will also give a resonance with to2 = 103 — toi, k2 = k3 — ki,
and similarly 102, k2 and C03, k3 lead to to\, ki.
In particular, for an isotropic body to and k are related by to = ctk or
to = cik, with ci > ct. It is easy to see in which cases either of these relations
can hold for each of the three combinations
wi, ki; 0)2, k2; a> 3 = CO1+C02, k3 = ki + k2.
If ki and k2 are not in the same direction, A3 < ki + &2, and so it is clear that
resonance can then occur only in the following two cases: (1) the waves with
toi, ki and 002, k2 are transverse and that with C03, k3 longitudinal ; (2) one of
the waves with toi, ki and a>2, k2 is transverse and the other longitudinal, and
that with a>3, k3 is longitudinal. If the vectors ki and k2 are in the same
direction, however, resonance is possible when all three waves are longi-
tudinal or all three are transverse.
The anharmonic effect involving resonance occurs not only when several
monochromatic waves are superposed, but also when there is only one wave,
§26 Anharmonic vibrations 121
with parameters a>i, ki. In this case the right-hand sides of the equations of
motion contain terms proportional to e 2 ^ k i- r-0, i () . if a)1 and ki satisfy the
usual condition, however, then 2o>i and 2ki do so too, since this condition is
homogeneous and of degree one. Thus the anharmonic effect results in the
appearance, besides the monochromatic waves with o>i, ki previously ob-
tained, of waves with 2oji, 2ki, i.e. with twice the frequency and twice the
wave vector, and amplitude increasing with time.
Finally, we may briefly discuss how we can set up the equations of motion,
allowing for the anharmonic terms. The strain tensor must now be given
by the complete expression (1.3):
_\(dui eto* dui dui \
2\dxjc dxi dxi dxjeJ*
in which the terms quadratic in ut can not be neglected. Next, the general
expression for the energy densityf <?, in bodies having a given symmetry,
must be written as a scalar formed from the components of the tensor um
and some constant tensors characteristic of the substance involved; this
scalar will contain terms up to a given power of tint. Substituting the ex-
pression (26.1) for Uik and omitting terms in m of higher orders than that
power, we find the energy $ as a function of the derivatives dui/dxjc to the
required accuracy.
In order to obtain the equations of motion, we notice the following result.
The variation 8$ may be written
S^ = 8—,
d(duijdx]c) dxk
or, putting
d&
<*ik — ; » (26.2)
d(dui/dxk)
dStii d dcrijc
o& = a ilc — = — — (ar ik Sui) — 8ui— .
dxjc 0X]c oxjc
The coefficients of — 8ut are the components of the force per unit volume of
the body. They formally appear the same as before, and so the equations of
motion can again be written
poiii = doujdxjc, (26.3)
f We here use the internal energy S, and not the free energy F, since adiabatic vibrations are
involved.
5
122 Elastic Waves §26
where po is the density of the undeformed body, and the components of the
tensor atk are now given by (26.2), with g correct to the required accuracy.
The tensor ow is no longer symmetrical, f
PROBLEM
Write down the general expression for the elastic energy of an isotropic body in the third
approximation.
Solution. From the components of a symmetrical tensor of rank two we can form two
quadratic scalars (urn* and «n 2 ) and three cubic scalars («u 8 , whw** 8 and uucUnujci). Hence the
most general scalar containing terms quadratic and cubic in wie, with scalar coefficients
(since the body is isotropic), is
6 = pUiic 2 + {$K-$fi)uu 2 + $Aui k Uiiu k i+Bui k 2 uii + lCuuZ;
the coefficients of utk* and tin* have been expressed in terms of the moduli of compression
and rigidity, and A, B, C are three new constants. Substituting the expression (26.1) for
in* and retaining terms up to and including the third order, we find the elastic energy to be
dui dujA 2 v/^ M A 2
\dxic Cxi I \dxil
+
2
dui dui dui dui /dui\''
OXjc OXi OXjc OXi \OXjcJ
du t du k dui ,dui du k du t ,^/3«A 3
+h A +iB +$c — .
dx k dxi dxi dx k dxt dxi \dxiJ
f It should be emphasised that <7<fc is no longer the momentum flux density (the stress tensor).
In the ordinary theory this interpretation was derived by integrating the body force density
doikjdxic over the volume of the body. This derivation depended on the fact that, in performing the
integration, we made no distinction between the co-ordinates of points in the body before and after
the deformation. In subsequent approximations, however, this distinction must be made, and the
surface bounding the region of integration is not the same as the actual surface of the region considered
after the deformation.
It has been shown in §2 that the symmetry of the tensor am is due to the conservation of angular
momentum. This result no longer holds, since the angular momentum density is not Xiiijc — **m<
but (Xi+ujuk -(**+«*)«<•
CHAPTER IV
DISLOCATIONS!
§27. Elastic deformations in the presence of a dislocation
Elastic deformations in a crystal may arise not only by the action of external
forces on it but also because of internal structural defects present in the crystal.
The principal type of defect that influences the mechanical properties of cry-
stals is called a dislocation. The study of the properties of dislocations on the
atomic or microscopic scale is not, of course, within the scope of this book ; we
shall here consider only purely macroscopic aspects of the phenomenon as it
affects elasticity theory. For a better understanding of the physical significance
of the relations obtained, however, we shall first give two simple examples to
show what is the nature of dislocation defects as regards the structure of the
crystal lattice.
Let us imagine that an "extra" half-plane is put into a crystal lattice of
which a cross-section is shown in Fig. 22 ; in this diagram, the added half-plane
Fig. 22
is the upper half of the _y#-plane. The edge of this half -plane (the #-axis, at
right angles to the plane of the diagram) is then called an edge dislocation. In
the immediate neighbourhood of the dislocation the crystal lattice is greatly
distorted, but even at a distance of a few lattice periods the crystal planes fit
together in an almost regular manner. The deformation nevertheless exists
even far from the dislocation. It is clearly seen on going round a closed circuit
of lattice points in the xy-plane, with the origin within the circuit: if the
| This chapter was written jointly with A. M. Kosevich.
123
124
Dislocations
§27
displacement of each point from its position intheideal lattice is denoted by the
vector u, the total increment of this vector around the circuit will not be zero,
but equals one lattice period in the x-direction.
Another type of dislocation may be visualised as the result of "cutting" the
lattice along a half-plane and then shifting the parts of the lattice on either
side of the cut in opposite directions to a distance of one lattice period parallel
to the edge of the cut (then called a screw dislocation). Such a dislocation
converts the lattice planes into a helicoidal surface, like a spiral staircase
without the steps. In a complete circuit round the dislocation line (the axis of
the helicoidal surface) the lattice point displacement vector increment is one
lattice period along that axis. Figure 23 shows a diagram of such a cut.
Fig. 23
Microscopically, a dislocation deformation of a crystal regarded as a
continuous medium has the following general property; after a passage round
any closed contour L which encloses the dislocation line D, the elastic
displacement vector u receives a certain finite increment b which is equal to
one of the lattice vectors in magnitude and direction; the constant vector b is
called the Burgers vector of the dislocation concerned. This property may be
•expressed as
)6ui = q) dx/c = — bf, (27.1)
J dxje
where the direction in which the contour is traversed and the chosen direction
of the tangent vector t to the dislocation line are assumed to be related by the
corkscrew rulef (Fig. 24). The dislocation line itself is a line of singularities of
the deformation field.
It is evident that the Burgers vector b is necessarily constant along the dis-
location line, and also that this line cannot simply terminate within the crystal:
it must either reach the surface of the crystal at both ends or (as usually hap-
pens in actual cases) form a closed loop.
f The simple cases of edge and screw dislocations mentioned above correspond to straight lines D
-with x J_ b and t || b. We may also note that in the representation given by Fig. 22 edge dislocations
-with opposite directions of b differ in that the "extra" crystal half-plane lies above or below the xz-
plane; such dislocations are said to have opposite signs.
§27 Elastic deformations in the presence of a dislocation 125
The condition (27.1) signifies, therefore, that in the presence of a dislocation
the displacement vector is not a single-valued function of the co-ordinates,
but receives a certain increment in a passage round the dislocation line.
Physically, of course, there is no ambiguity: the increment b denotes an addi-
tional displacement of the lattice points equal to a lattice vector, and this does
not affect the lattice itself.
In the subsequent discussion it is convenient to use the notation
w>ik — dujcjdxi, (27.2)
so that the condition (27.1) becomes
I
WW dxt = - bk. (27.3)
The (unsymmetrical) tensor wm is called the distortion tensor. Its symmetrical
part gives the ordinary strain tensor:
uik = l(mk + mi)- (27.4)
According to the foregoing discussion the tensors zvtk and utk, and therefore
the stress tensor o-^, are single-valued functions of the co-ordinates, unlike
the function u(r).
The condition (27.3) may also be written in a differential form. To do so,
we transform the integral round the contour L into one over a surface Sl
spanning this contour :f
r C 8w mk
<bzv m jc ax m = earn— - — aft-
v »/ C/XT
dwmk
■Urn
~L St.
dxi
The constant vector bk is written as an integral over the same surface by
t The transformation is made, according to Stokes' theorem, by replacing dx m by the operator
d/ietim djdxi, where eam is the antisymmetric unit tensor.
126 Dislocations §27
means of the two-dimensional delta function :
h= (nhSftdfi, (27.5)
i
where \ is the two-dimensional radius vector taken from the axis of the dis-
location in the plane perpendicular to the vector t at the point considered.
Since the contour L is arbitrary, the integrals can be equal only if the inte-
grands are equal:
eumdwmkldxi= -TihSfe). (27.6)
This is the required differential form.f
The displacement field u(r) around the dislocation can be expressed in a
general form if we know the Green's tensor G^(r) of the equations of equilib-
rium of the anisotropic medium considered, i.e. the function which determines
the displacement component Ui produced in an infinite medium by a unit
force applied at the origin along the a^-axis (see §8). This can easily be done
by using the following formal device.
Instead of seeking many-valued solutions of the equations of equilibrium,
we shall regard u(r) as a single-valued function, which undergoes a fixed
discontinuity b on some arbitrarily chosen surface Sd spanning the dislocation
loop D. Then the strain tensor formally defined by (27.4) will have a delta-
function singularity on the "surface of discontinuity":
Wt™ = l{nib k + n k bi)8(Q, (27.7)
where £ is a co-ordinate measured from the surf ace Sd along the normal n
(which is in the direction relative to t shown in Fig. 24).
Since there is no actual physical singularity in the space around the dis-
location, the stress tensor ow must, as already mentioned, be a single-valued
and everywhere continuous function. The strain tensor (27.7), however, is
formally related to a stress tensor anc^ = Xncim uimS s \ which also has a
singularity on the surface Sd- In order to eliminate this we must define ficti-
tious body forces distributed over the surface Sd with a certain density f( s K The
equations of equilibrium in the presence of body forces are doijc/dxjc +ft^ =
(cf. (2.7)). Hence it is clear that we must put
ft*) = JL_ = _ X mm -^—. (27.8)
OXjc CXjc
Thus the problem of finding the many-valued function u(r) is equivalent to
that of finding a single-valued but discontinuous function in the presence of
f To avoid misunderstanding it should be noted that on the dislocation line itself (( -*■ 0), which is
a line of singularities, the representation of the ttHk as the derivatives (27.2) is no longer meaningful.
§27 Elastic deformations in the presence of a dislocation 127
body forces given by formulae (27.7) and (27.8). We can now use the formula
^(r) = J^(r-r')/^)(r')dF.
We substitute (27.8) and integrate by parts; the integration with the delta
function is then trivial, giving
C d
u t (r) = - Xjumbm « r — G*y(r - r')d/'. (27.9)
J OXlc
This solves the problem. -f
The deformation (27.9) has its simplest form far from the closed dislocation
loop. If we imagine the loop to be situated near the origin, then at distances r
large compared with the linear dimensions of the loop we have
u t {t) = -Xjkimdi m dGij(r)jdx k , (27.10)
where
dm = Sib k , Si = n t df= le m i>x k dxi, (27.11)
s D d
and enci is the antisymmetric unit tensor. The axial vector S has components
equal to the areas bounded by the projections of the loop D on planes perpen-
dicular to the corresponding co-ordinate axes ; the tensor due may be called the
dislocation moment tensor. The components of the tensor Gy are first-order
homogeneous functions of the co-ordinates x, y, z (see §8, Problem). We
therefore see from (27.10) that Ui~\/r 2 , and the corresponding stress field
It is also easy to ascertain the way in which the elastic stresses vary with
distance near a straight dislocation. In cylindrical polar co-ordinates z, r, <f>
(with the s-axis along the dislocation line) the deformation will depend only
on r and <f>. The integral (27.3) must, in particular, be unchanged by an
arbitrary change in the size of any contour in the xy-plane which leaves the
shape of the contour the same. It is clear that this can be true only if all the
woc^X jr. The tensor w^, and therefore the stresses cr^, will be proportional
to the same power, 1/r.J
f The tensor Gn for an anisotropic medium has been derived in the paper by I. M. Lifshitz and
L. N. RozentsveIg quoted in §8, Problem. This tensor is in general very complicated. For a straight
dislocation, which corresponds to a two-dimensional problem of elasticity theory, it may be simpler to
solve the equations of equilibrium directly.
% Attention is drawn to a certain analogy between the elastic deformation field round a dislocation
line and the magnetic field of constant line currents. The current is replaced by the Burgers vector,
which must be constant along the dislocation line, like the current. Similar analogies will also be readily
seen in the relations given below. However, quite apart from the entirely different nature of the two
physical effects, these analogies are not far-reaching, because the tensor character of the corresponding
quantities is different.
128 Dislocations §27
Although we have hitherto spoken only of dislocations, the formulae de-
rived are applicable also to deformations caused by other kinds of defect in
the crystal structure. Dislocations are linear defects; there exist also defects in
which the regular structure is interrupted through a region near a given
surface. | Such a defect can be macroscopically described as a surface of dis-
continuity on which the displacement vector u is discontinuous but the stresses
aik are continuous, by virtue of the equilibrium conditions. If the discontinuity
b is the same everywhere on the surface, the resulting strain is just the same as
that due to a dislocation along the edge of the surface. The only difference is
that the vector b is not equal to a lattice vector. However, the position of the
surface Sd discussed above is no longer arbitrary; it must coincide with the
actual physical discontinuity. Such a surface of discontinuity involves a certain
additional energy which may be described by means of an appropriate surface-
tension coefficient.
PROBLEMS
Problem 1 . Derive the differential equations of equilibrium for a dislocation deformation
in an isotropic medium, expressed in terms of the displacement vector.^
Solution. In terms of the stress tensor or strain tensor the equations of equilibrium have the
usual form daikjdxu = or, substituting a i]c from (5.11),
duM a du n
dxjc l—2a dxi
To convert to the vector u we must use the differential condition (27.6). Multiplying (27.6)
by even and summing over * and k, we obtain§
dWnk dwjcjc
= -(TXb)»8(|). (2)
Writing (1) in the form
dxjc dx n
2 o ' 2~ ~ l"~ Z Z = "
dx^ dxjc 1 — 2ct dx t
and substituting (2), we find
dzvjci 1 dzon
dxjc l—2a dxi
Now changing to u in accordance with (27.2), we find the required equation for the multi-
valued function u(r) :
1
Au+— — - grad div u = xXbS(^). (3)
1 — La
t A well-known example of a defect of this type is a narrow twinned layer in a crystal.
\ The physical meaning of this and other problems relating to an isotropic medium is purely
conventional, since actual dislocations by their nature occur only in crystals, i.e. in anisotropic media.
Such problems have illustrative value, however.
§ Using also the formula eumetkn = Si k S mn — Sj„fi mfc .
§27 Elastic deformations in the presence of a dislocation 129
Problem 2. Determine the deformation near a straight screw dislocation in an isotropic
medium.
Solution. We take cylindrical polar co-ordinates z, r, ^, with the sr-axis along the disloca-
tion line; the Burgers vector is b x = b y = 0, b z = b. It is evident from symmetry that the
displacement u is parallel to the #-axis and is independent of the co-ordinate z. The equation
of equilibrium (3), Problem 1, reduces to A«z = 0. The solution which satisfies the condition
(27.1) isf u z = b<f>j2ir. The only non-zero components of the tensors utk and 0% are m z <j> =
bjAirr, o z $ = fiil2irr, and the deformation is therefore a pure shear.
The free energy of the dislocation (per unit length) is given by the integral
fib 2 rdr
4tt J r '
which diverges logarithmically at both limits. As the lower limit we must take the order of
magnitude of the interatomic distances (~b), at which the deformation is large and the macro-
scopic theory is inapplicable. The upper limit is determined by a dimension of the order of the
length L of the dislocation. Then F = (\tb 2 \A-n) log (L/b). The energy of the deformation in the
"core' ' of the dislocation near its axis (in a region of cross-sectional area ~ b 2 ) can be estimated
as ~ fib 2 . When log (Ljb) ^> 1 this energy is small in comparison with that of the elastic
deformation field. %
Problem 3. Determine the internal stresses in an anisotropic medium near a screw disloca-
tion which is perpendicular to a plane of symmetry of the crystal.
Solution. We take co-ordinates x, y, z so that the #-axis is along the dislocation line, and
again write b z = b. The vector u again has only the component u z = u(x,y). Since the xy-
plane is a plane of symmetry, all the components of the tensor Xikim are zero which contain the
suffix z an odd number of times. Thus only two components of the tensor aw are non-zero :
o X z — A-xzxz~^ V^xzyz
du du
- — \-^xzyz~ ,
ox cy
du du
Vyz — Ayzxz-7- + Ayzyz~z~-
ox cy
We define a two-dimensional vector a and a two-dimensional tensor A a g: <r a = o az , A a $ =
A<x«3 3 (a = 1, 2). Then <r a = \ a &8ul8x$, and the equation of equilibrium becomes div o = 0.
The required solution of this equation must satisfy the condition (27.1) : $ grad u • dl = b.
In this form, the problem is the same as that of finding the magnetic induction and magnetic
field (represented by o and grad u) in an anisotropic medium of magnetic permeability A a p
near a straight current of strength I = cb\\-n. Using the solution derived in electrodynamics,
we obtain§
o\,, = —
b KftepyzXy
2n VH-*«V*«'*/ '
where |A| is the determinant of the tensor A a p.
t In all the problems on straight dislocations we take the vector t in the negative ar-direction.
t These estimates are general ones and are valid in order of magnitude for any dislocation (and not
only for a screw dislocation).
It should be noted that in practice the values of log (L/b) are usually not very large, and the energy
of the "core" is therefore a considerable fraction of the total energy of the dislocation.
§ See Electrodynamics of Continuous Media, §29, Problem 5.
130 Dislocations §27
Problem 4. Determine the deformation near a straight edge dislocation in an isotropic
medium.
Solution. Let the s-axis be along the dislocation line, and the Burgers vector be b x = b,
by = b z = 0. It is evident from the symmetry of the problem that the displacement vector
lies in the xy-plane and is independent of z, so that the problem is a two-dimensional one. In
the rest of this solution all vectors and vector operations are two-dimensional in the xy-plane.
We shall seek a solution of the equation
AuH grad div u = -&jS(r)
1 — 2ct
(see Problem 1 ; j is a unit vector along the j>-axis) in the form u = u(°> + w, where u<°> is a
vector with components u^ x = b<l>l2ir, w<% = (fc/2-n-) log r; these are the imaginary and real
parts of (6/2tt) log (x+iy), r and <f> being polar co-ordinates in the xy-plane. This vector
satisfies the condition (27.1). The problem therefore reduces to finding the single- valued
function w. Since, as is easily verified, div u<°> = 0, Au<°> = £j8(r), it follows that w satisfies
the equation
Aw + grad div w = -2£j8(r).
1 — 2cr
This is the equation of equilibrium under forces concentrated along the sr-axis with volume
density £6jS(r)/2(l + a) ; cf. §8, Problem, equation (1). By means of the Green's tensor found
in that problem for an infinite medium, the calculation of w is reduced to that of the integral
W =
a) J [ r m
r(l-a)
u
R = ^2 + ^2).
The result is
b j y 1 xy
u x = — — {tan -1 --r-
2tt 1 x 2(1 — a) x 2 +y 2 j
b i 1-2<t 1 x 2 |
The stress tensor calculated from this has Cartesian components
y(3x 2 +y 2 )
vxx = - bD
Oyy = bD
o X y = bD
(x 2 +y 2 ) 2
y(x 2 — y 2 )
(x 2 +y 2 ) 2
x(x 2 — y 2 )
(x 2 +y 2 ) 2
and polar components
where D = /*/27r(l — a).
<*rr = o H = -(bD/r) sin <f>,
a r4> = (bDjr) cos <f>,
§28 The action of a stress field on a dislocation 131
Problem 5. An infinity of identical parallel straight edge dislocations in an isotropic
medium lie in one plane perpendicular to their Burgers vectors and at equal distances h apart.
Find the shear stresses due to such a "dislocation wall" at distances large compared with h.
Solution. Let the dislocations be in the ys-plane and parallel to the sr-axis. According to
the results of Problem 4, the total stress due to all the dislocations at the point (x, y) is given by
the sum
a xy (x, y) = bDx> — — — -.
This may be written in the form
a xy
a
= -bD-
h
■ 3/(«,0)
7(a, p) + a-
da.
where
' + {^-nf
n = — °° v
According to Poisson's summation formula
00 00 00
Y/(«) = V j f( x )e 2 " ikx dx,
k = — °° — °°
find
d£ ^ . re****idg
f df ^ re*
77 Z7T
77 Z.77 -«r— v
= -+ > e-2^^«cos 27rfyS.
a a ^
ft = i
When a = »/A^> 1 only the first term need be retained in the sum over k, and the result is
bx
a xy = ^ 7T 2]T > _ e -27rx/h C0S(277}>//*).
h 2
Thus the stresses decrease exponentially away from the wall.
§28. The action of a stress field on a dislocation
Let us consider a dislocation loop D in a field of elastic stresses a\^
created by given external loads, and calculate the force on the loop in such a
field.
According to the general rules, this must be done by finding the work 8R
done by internal stresses in an infinitesimal displacement of the loop D. If
1 32 Dislocations §28
Sttjfc is the change in the strain tensor due to this displacement, we have from
(3.1)f
sr = - o ik mui k dv.
~
dujc
* -dV
Since the distribution of the stresses o^ e) is assumed independent of the
position of the dislocation, we can take the difference symbol S outside the
integral. Using also the symmetry of the tensor ot k < e ) and the equation of
equilibrium dai k ^\dx k = 0, we can write
8R = -8L ik ^u ik dV
J CXi
= -h[- {a ik ^u k )dV. (28.1)
J OXf
As explained in §27, we shall regard the displacement u as a single-valued
function having a discontinuity on some surface Sd spanning the line D. Then
the volume integral in (28.1) can be transformed into an integral over a closed
surface consisting of the upper and lower surfaces of the cut Sd, joined by a
tubular lateral surface of infinitesimal width enclosing the line D. The values
of the continuous quantities <Ji k ^ are the same on both surfaces, but the values
of u differ by a given amount b. We therefore obtain^
jcr^dfi. (28.2)
8R = -b k 8
x
Let each element of length dl of the dislocation be displaced by an amount
Sr. This displacement causes a change in the area of the surface Sd, the
elementary change being Sf = Sr X dl, i.e. 8fi = ei mn 8x m dl n = ei mn 8x m r n dl.
The work (28.2) therefore becomes a line integral round the dislocation loop :
8R = — Sb k ei mn cr k i^8x m rndl,
where t is the tangent vector to D.
The coefficient of 8x m in the integrand is minus the force f m on unit length
f To avoid misunderstanding we must emphasise that Suae in this formula is to be taken (in accordance
with the sense of this quantity in (3.1)) as the total (geometrical) change in the deformation following an
infinitesimal movement of the dislocation. It comprises both elastic and plastic (see §29) parts.
\ The integral over the tubular lateral surface of radius p vanishes as p — ► 0, since the uie become
infinite more slowly than 1/p.
§28 The action of a stress field on a dislocation 133
of the dislocation line. Thus
fi = e iklTk°lrn e) bm (28.3)
(M. Peach and J. S. Koehler 1950). We may note that the force f is perpen-
dicular to the vector t, i.e. to the dislocation line, and also to the vector
oik (e) bk-
The plane which is defined by the vectors t and b at each point of the dis-
location is called the slip plane of the corresponding element of the dislocation ;
for every element this plane of course touches the slip surface of the whole
dislocation, which is a cylindrical surface with generators parallel to the Bur-
gers vector b of the dislocation. The distinctive physical property of the slip
plane is that it is the only one in which a comparatively easy mechanical dis-
placement of the dislocation is possible.-f- For this reason it is of interest to
determine the force (28.3) on this plane.
Let k be a vector normal to the dislocation line in the slip plane. Then the
required force component (f ±t say) is f ± = Kifa = eikiKiT k b m ai m < e \ or
f x = vtomPbm, (28.4)
where v = k Xt is a vector normal to the slip plane. Since the vectors b and v
are perpendicular, we see that the force f x is determined by only one compo-
nent oimS e ) if two of the co-ordinate axes are taken along these vectors.
The total force acting on the whole dislocation loop is
Fi = encib m ixri m {e) dx k . (28.5)
This is zero except for a non-uniform stress field ; when a\ m ^ = constant,
the integral is §dx k = 0. If the stress field varies only slightly over the loop,
we can write
ri = enab m — (pa^d^A;,
P D
the loop being regarded as situated near the origin .This force can be expressed
in terms of the dislocation moment du defined by (27.11):
F t = d k idad e) ldxi. (28.6)
PROBLEMS
Problem 1. Find the force of interaction between two parallel screw dislocations in an
isotropic medium.
f This fact follows from the microscopic form of a dislocation defect. For example, to move the
edge dislocation shown in Fig. 22 in its slip plane (the xs-plane) comparatively slight movements of the
atoms are sufficient, which make crystal planes farther and farther from the yar-plane (but still parallel to
it) into "extra" planes.
The movement of the dislocation in other directions can occur only by diffusion processes. For
example, the dislocation shown in Fig. 22 can move in the yar-plane only when atoms leave the "extra"
half-plane by diffusion. Such a process can be of practical importance only at fairly high temperatures.
134 Dislocations §29
Solution. The force per unit length acting on one dislocation in the stress field due to the
other dislocation is determined from formula (28.4), using the results of §27, Problem 2. It is
a radial force of magnitude/ = [ib\b 2 \2m : Dislocations of like sign (b x b 2 > 0) repel, while
those of unlike sign (bib 2 < 0) attract.
Problem 2. A straight screw dislocation lies parallel to the plane free surface of an iso-
tropic medium. Find the force acting on the dislocation.
Solution. Let the yz-plane be the surface of the body, and let the dislocation be parallel to
the .sr-axis with co-ordinates * = xo, y — 0.
The stress field which leaves the surface of the medium a free surface is described by the sum
of the fields of the dislocation and its image in the y^-plane, considered to lie in an infinite
medium :
fxb
y y
[ib
Gyz = —
2ttL(x— xo) 2 +y 2 (x + xo) 2 +y 2
X — Xq X + Xq
2tt L (x — xo) 2 +y 2 (x+xo) 2 +y 2 _
Such a field exerts a force on the dislocation considered which is equal to the attraction exerted
by its image, i.e. the dislocation is attracted to the surface of the medium by a force
/ = nb'ftnxo.
Problem 3. Find the force of interaction between two parallel edge dislocations in an
isotropic medium which are in parallel slip planes.
Solution. Let the slip planes be parallel to the xsr-plane and let the ar-axis be parallel to the
dislocation lines ; as in §27, Problem 4, we put t s = — l,b x = b. Then the force on unit length
of the dislocation in the field of elastic stresses ow has components f x = bo xy ,f y = — bo xx .
In the case considered, oik is determined by the expressions derived in §27, Problem 4. If one
dislocation is along the z-axis, it exerts on the other dislocation (passing through the point
(x, y, 0)) a force whose polar components are f r = b\b 2 Djr, f<\> = {bib 2 Djr) sin 2<f>, D =
nl2ir(l — a). The component of this force in the slip plane is f x = (bib 2 Dlr) cos <f> cos 2<j>,
which is zero when <f> = \-n or \tt. The former position corresponds to stable equilibrium when
6i&2 > 0» the latter when bib 2 < 0.
§29. A continuous distribution of dislocations
If a crystal contains several dislocations at the same time which are at
relatively short distances apart (although far apart compared with the lattice
constant, of course), it is useful to treat them by means of an averaging process :
we consider "physically infinitesimal" volume elements in the crystal with a
large number of dislocation lines through each.
An equation which expressed a fundamental property of dislocation deforma-
tions can be formulated by a natural generalisation of equation (27.6). We
define a tensor pw (the dislocation density tensor) such that its integral over a
surface spanning any contour L is equal to the sum b of the Burgers vectors
of all the dislocation lines embraced by the contour :
f.
PiJcdft = b k . (29.1)
The continuous functions pw describe the distribution of dislocations in the
§29 A continuous distribution of dislocations 135
crystal. This tensor now replaces the expression on the right of equation
(27.6):
eumdWmk/ %*% = - pik- (29.2)
This equation shows that the tensor p ik must satisfy the condition
8paldx t = Q; (29.3)
for a single dislocation, this condition simply states that the Burgers vector is
constant along the dislocation line.
When the dislocations are treated in this way, the tensor w ik becomes a
primary quantity describing the deformation and determining the strain ten-
sor through (27.4). A displacement vector u related to w i1c by the definition
(27.2) cannot exist; this is clear from the fact that with such a definition the
left-hand side of equation (29.2) would be identically zero throughout the
crystal.
So far we have assumed the dislocations to be at rest. Let us now see how a
set of equations may be formulated so as to allow in principle elastic deforma-
tions and stresses in a medium where dislocations are moving in a given man-
nerf (E. Kroner and G. Rieder 1956).
Equation (29.2) is independent of whether the dislocations are at rest or in
motion. The tensor w ilc still determines the elastic deformation; its symmetri-
cal part is the elastic strain tensor, which is related to the stress tensor in the
usual way, by Hooke's law.
This equation, however, is now insufficient for a complete formulation of
the problem. The full set of equations must also determine the velocity v of
the points in the medium.
It must be borne in mind that the movement of dislocations causes not only
a change in the elastic deformation but also a change in the shape of the crystal
which does not involve stresses, i.e. aplastic deformation. The motion of dis-
locations is in fact a mechanism of plastic deformation. This is clearly illustra-
ted by Fig. 25, where the passage of the edge dislocation from left to right
causes the part of the crystal above the slip plane to be shifted to the right by
one lattice period; since the lattice is then regular, the crystal remains un-
stressed. Unlike an elastic deformation, which is uniquely defined by the
thermodynamic state of the body, a plastic deformation depends on the process
which occurs. In considering dislocations at rest we have no need to distin-
guish elastic and plastic deformations, since we are concerned only with
stresses which are independent of the previous history of the crystal.
Let u be the geometrical displacement vector of points in the medium,
measured, say, from their position before the deformation process begins; its
f We shall not discuss here the problem of determining this motion itself from the forces applied to
the body. The solution of such a problem requires a detailed study of the microscopic mechanism of the
motion of dislocations and their retardation by various defects, which must take account of the
conditions occurring in actual crystals.
136
Dislocations
§29
time derivative ii = v. If the "total distortion" tensor W ik = dut/dxt is
formed from the vector u, its "plastic part" «^< pl ) is obtained by subtracting
Fig. 25
from Wik the "elastic distortion" tensor, which is the same as the tensor zcijc in
(29.2). We use the notation
-ja=fai*Wlto; (29.4)
the symmetrical part ofjw gives the rate of variation of the plastic deformation
tensor: the change in e^& (pl) in an infinitesimal time interval 8t is
Si^CPD = -ft/tt+AOS*. (29.5)
We may note, in particular, that, if a plastic deformation occurs without
destroying the continuity of the body, the trace of the tensor jm is zero : a
plastic deformation causes no extension or compression of the body (which
would always involve the appearance of internal stresses), i.e. u^^ = 0, and
therefore;^ = -du kk W/dt = 0.
§29 A continuous distribution of dislocations 137
Substituting in the definition (29.4) Wik^ = Wuc — ww, we can write it as
= -r-+ja, (29. f^
dt dxt
an equation which relates the rates of change of the elastic and plastic deforma-
tions. Here the /^ must be regarded as given quantities which must satisfy
conditions ensuring the compatibility of equations (29.6) and (29.2). These
conditions are found by differentiating (29.2) with respect to time and sub-
stituting (29.6), and are
^+^ = 0. (29.7)
Ot OX\
The complete set of equations is given by (29.2) and (29.6), together with the
dynamical equations
pvi = datjcldxjc, (29.8)
where a ilc = XtjcimUim = At^m^zm-The tensors puc and fa which appear in
these equations are given functions of the co-ordinates (and time) which
describe the distribution and movement of the dislocations. These functions
must satisfy the compatibility conditions of equations (29.2) with one another
and with (29.6), which are given by (29.3) and (29.7).
The condition (29.7) may be regarded as a differential expression of the "law
of conservation of the Burgers vector" in the medium: integrating both sides
of this equation over a surface spanning some closed line L, defining by (29.1)
the total Burgers vector b of the dislocations embraced by L, and using
Stokes' theorem, we obtain
~df
= -jjikdxi. (29.9)
The form of this equation shows that the integral on the right gives the "flux"
of the Burgers vector through the contour L per unit time, i.e. the Burgers
vector carried across L by moving dislocations. We may therefore caliy^ the
dislocation flux density tensor.
In particular, it is clear that for an isolated dislocation loop the tensor jw has
the form
jik = eumpucVm
= e am TiV m b k o{l), (29.10)
puc being given by (27.6), and V being the velocity of the dislocation line at a
particular point on it. The flux vector through the element dl of the contour
L is jijcdli and is proportional to dl^x XV = V«dl Xt, i.e. the component of V
in a direction perpendicular to both dl and t ; from geometrical considerations
6
138 Dislocations §29
it is evident that this is correct, since only that velocity component causes the
dislocation to intersect the element dl.
We may note that the trace of the tensor (29.10) is proportional to the com-
ponent of the velocity of the dislocation along the normal to its slip plane. It
has been mentioned above that the absence of any inelastic change in density
of the medium is ensured by the condition/^ = 0. We see that for an indivi-
dual dislocation this condition signifies motion in the slip plane, in accordance
with the previous discussion of the physical nature of the movement of dis-
locations; see the last footnote to §28.
Finally, let us consider the case where dislocation loops are distributed in
the crystal in such a way that their total Burgers vector (denoted by B) is zero.f
Fig. 26
This condition signifies that integration over any cross-section of the body
gives
pikdfi = 0.
(29.11)
From this it follows that the dislocation density in this case can be written as
pik = eumBPmkl dxi (29 . 1 2)
(F. Kroupa 1962); then the integral (29.11) becomes an integral along a con-
tour outside the body, and is zero. It may also be noted that the expression
(29.12) necessarily satisfies the condition (29.3).
It is easy to see that the tensor P^ thus defined represents the dislocation
moment density in the deformed crystal, and may therefore be called the
"dislocation polarisation": the total dislocation moment Afc of the crystal is,
by definition,
Dm — ^ Stbjc = leum£hi>xidx f
D
= J eamXlpmkdV,
t The presence of a dislocation involves a certain bending of the crystal, as shown diagrammatically
in Fig. 26 (greatly exaggerated). The condition B = means that there is no macroscopic bending of
the crystal as a whole.
§30 Distribution of interacting dislocations 139
where the summation is over all dislocation loops and the integration is over
the whole volume of the crystal. Substituting (29.12), we obtain
dP a ic
Dm = \ eu m e mpg xi— — dV
J OXp
-If.
uXi OXm/
and, after integrating by parts in each term,
D ik = jp ik dV. (29.13)
The dislocation flux density is given in terms of the same tensor Pi k by
jiJc=-dPi k l8L (29.14)
This is easily seen, for example, by calculating the integral jj\ k dV over an
arbitrary part of the volume of the body, using the expression (29.10), to give
a sum over all dislocation loops within that volume. We may note that the
expression (29.14) together with (29.12) automatically satisfies the condition
(29.7).
A comparison of (29.14) with (29.4) shows that Sto tk ^ = &P ik . If we
agree to regard the plastic deformation as absent in the state with Pm = 0,
then wifctoU = P^,f and
Wik = Wit-to*®) = dujcldxt-Pijc, (29.15)
where u k is again the vector of the total geometrical displacement from the
position in the undeformed state. Equation (29.6) is then satisfied identically,
and the dynamical equation (29.8) becomes
pUi-Xm m d 2 u m ldx k dxi = -X ik i m dPi m jdx k . (29.16)
Thus the determination of the elastic deformation due to moving dislocations
with B = reduces to a problem of ordinary elasticity theory with body forces
distributed in the crystal with density —hkimdPimfixk (A. M. Kosevich
1963).
§30. Distribution of interacting dislocations
Let us consider a large number of similar straight dislocations lying
parallel in the same slip plane, and derive an equation to determine their
equilibrium distribution. Let the sr-axis be parallel to the dislocations, and
the xs-plane be the slip plane.
t It is assumed that the entire deformation process occurs with B = 0. This point must be empha-
sised, since there is a fundamental difference between the tensors P ik and w Jfc (pl) : whereas P ik is a func-
tion of the state of the body, the tensor «.- Mr (pl> is not, but depends on the process which has brought the
body into that state.
140 Dislocations §30
We shall suppose for definiteness that the Burgers vectors of the disloca-
tions are in the ^-direction. Then the force in the slip plane on unit length of
a dislocation is ba xyy where a xy is the stress at the position of the dislocation.
The stresses created by one straight dislocation (and acting on another
dislocation) decrease inversely as the distance from it. The stress at a point x
due to a dislocation at a point x' is therefore bD/(x — x'), where D is a constant
of the order of the elastic moduli of the crystal. It may be shown that this
constant D is positive, i.e. two like dislocations in the same slip plane repel
each other.f
Let p(x) be the line density of dislocations on a segment (a±, a^) of the
x-axis ; p(x)dx is the sum of the Burgers vectors of dislocations passing through
points in the interval dx. Then the total stress at a point x on the x-axis due to
all the dislocations is given by the integral
a>
a xy {x)^-D[^p-^. (30.1)
J £ — X
For points in the segment («i, a?) this integral must be taken as a principal
value in order to exclude the physically meaningless action of a dislocation on
itself.
If the crystal is also subjected to a two-dimensional stress field a xy ^ e \x, y)
in the ry-plane, caused by given external loads, each dislocation will be sub-
jected to a force b(a xy +p(x)), where for brevity p{x) denotes o xy W(x, 0). The
condition of equilibrium is that this force should be zero: a xy +p — 0, i.e.
D ? />(£)<*£ p(x)
P — = — — = w(x), (30.2)
J tj — x D
ai
where P denotes, as usual, the principal value. This is an integral equation to
determine the equilibrium distribution p(x). It is a singular equation with a
Cauchy kernel.
The solution of such an equation is equivalent to a problem in the theory of
functions of a complex variable which may be formulated as follows.
Let Q,(z) denote a function defined throughout the complex #-plane (cut
from a± to a?) as the integral
Q(*) = J"!
''^^ (30.3)
«l
I"
Let Q. + (x) and Q~(x) denote the limiting values of Q(#) on the upper and lower
edges of the cut. They are equal to similar integrals along the segment («i, #2)
f For an isotropic medium this has been proved in §28, Problem 3.
§30 Distribution of interacting dislocations 141
with an indentation in the form of an infinitesimal semicircle below or above
the point z = x respectively, i.e.
a 2
a ±(*) = P mLl. + i^x). (30.4)
ax
If p(£) satisfies equation (30.2), the principal value of the integral is w(x), and
we therefore have
Q+(x) + Qr(x) = 2(v(x), (30.5)
Q+(x) - Q-(x) = 2iiT P {x). (30.6)
Thus the problem of solving equation (30.2) is equivalent to that of finding an
analytic function £l(z) with the property (30.5); p(x) is then given by (30.6).
The physical conditions of the problem in question also require that Q( oo) = ;
this follows because far from the dislocations (#-»+ oo) the stresses a X y must
be zero (by the definition (30.3), a xy {x) = — DQ.(x) outside the segment
(fli, «2)).
Let us first consider the case where there are no external stresses (j>(x) = 0),
and the dislocations are constrained by some obstacles (lattice defects) at the
ends of the segment (a\> a%). When co(x) = we have from (30.5) CI +(x) =
— Q.~(x), i.e. the function Q,(z) must change sign in a passage round each of
the points «i, a%. This condition is satisfied by any function of the form
P(z)
Q(z) = — , (30.7)
V[{<*2-z)(z-ai)]
where P(z) is a polynomial. The condition Q,(co) = means that we must
take P(z) = 1 (apart from a constant coefficient), so that
°<*> = -7i7 Tr ^ (30 ' 8)
y[{a 2 -z)(z-ai)]
The required function p(x) will, according to (30.6), have the same form. The
coefficient is determined from the condition
a*
j P (£)d£ = B, (30.9)
ay
where B is the sum of the Burgers vectors of all the dislocations, and so we have
P(*)= /r , B v rr- (30.10)
77 y [(«2 — x){x — a\) J
We see that the dislocations pile up towards the obstacles at the ends of the
142 Dislocations §30
segment, with density inversely proportional to the square root of the distance
from the obstacle. The stress outside the segment (ai, a 2 ) increases in the same
manner as the ends of the segment are approached, e.g. for x>a 2
BD
\/[(x-a 2 )(a 2 -ai)]
In other words, the concentration of dislocations at the boundary leads to a
stress concentration beyond the boundary.
Let us now suppose that under the same conditions (obstacles at the fixed
ends of the segment) there is also an external stress field p(x). Let Q.o(z)
denote a function of the form (30.7), and let us rewrite equation (30.5)
divided by Qo + = — Qo~ as
Q+(x) Q-(x) 2w(x)
C1q + (x) Q,o~(x) Qq + (x)
A comparison of this with (30.6) shows that
a*
Q(*) 1 f off) d|
«i
where P(z) is a polynomial. A solution which satisfies the condition D( oo)
= is obtained by taking as Qo(#) the function (30.8) and putting P(z) = C>
a constant. The required function p(x) is hence found by means of (30.6), and
the result is
u>2
M— ,V[(^-l)(.- ai )] i 'J a ' tf)V[(fl2 - f)tf - ai) ^ +
+ . (30.12)
V[( a 2-x)(x-a{)]
The constant C is determined by the condition (30.9). Here also p(x) increases
as (^2 — x)~ 1/2 when x-^a 2 (and similarly when x->ai), and a similar concentra-
tion of stresses occurs on the other side of the boundary.
If there is an obstacle only on one side (at a 2 , say) the required solution
must satisfy the condition of finite stress for all x<a 2 , including the point x=
a\ ; the position of the latter point is not known beforehand and must be deter-
mined by solving the problem. With respect to £i{z) this means that Q.(a{)
must be finite. Such a function (satisfying also the condition Q(oo) = 0) is
obtained from the same formula (30.11) by taking for Q,q(z) the function
§30 Distribution of interacting dislocations 143
\Z[(z-ai)l(a 2 -z)], which is also of the form (30.7), and putting P(z) = in
(30.11). The result is
^ )= _I/^p]/^4^. (30.13)
«i
When x^ai, p(x) tends to zero as <\/(x-ai). The total stress a xy (x)+p(x)
tends to zero according to a similar law on the other side of the point a\.
Finally, let there be no obstacle at either end of the segment, and let the
dislocations be constrained only by external stresses />(#). The corresponding
Q(*) is obtained by putting in (30.11) Cl (z) = V[(«2-*)(*-«i)], P( z ) = °-
The condition ii(oo) = 0, however, here requires the fulfilment of a further
condition: taking the limit as z~>co in (30.11), we find
a 2
i
(30.14)
y[(«s-£)(£-«i)]
The function p(x) is given by
*)- 4^-^- ai)]p J v^-f)«-^] p? (30 - 15)
tti
the co-ordinates «i and «2 of the ends of the segment being determined by the
conditions (30.9) and (30.14).
PROBLEM
Find the distribution of dislocations in a uniform stress field p(x) = po over a segment
with obstacles at one or both ends.
Solution. When there is an obstacle at one end (a 2 ) the calculation of the integral (30.13)
gives
p0 IX- fli
" w - sy
«2 — X
The condition (30.9) determines the length of the segment occupied by dislocations :a 2 — a\ =
IBDjPo. Beyond the obstacle there is a concentration of stresses near it according to
/ A2 - « 1
<Txy=p0 •
V x — a<i
For a segment of length 2L bounded by two obstacles we take the origin of * at the midpoint
and obtain from (30.12)
pfx) = l—x + b) .
HK } 77a/(L2-*2) \ D J
144 Dislocations §31
§31. Equilibrium of a crack in an elastic medium
The problem of the equilibrium of a crack is somewhat distinctive among
the problems of elasticity theory. From the point of view of that theory, a
crack is a cavity in an elastic medium, which exists when internal stresses are
present in the medium and closes up when the load is removed. The shape and
size of the crack depend considerably on the stresses acting on it. The mathe-
matical feature of the problem is therefore that the boundary conditions are
given on a surface which is initially unknown and must itself be determined in
solving the problem.f
Let us consider a crack in an isotropic medium, of infinite length and uni-
form in the ^-direction and in a plane stress field vik {e) (x, y) ; this is a two-
dimensional problem of elasticity theory. We shall suppose that the stresses
are symmetrical about the centre of the cross-section of the crack. Then the
outline of the cross-section will also be symmetrical (Fig. 27). Let its length be
2L and its variable width h(x) ; since the crack is symmetrical, h( — x) = h(x)
We shall assume the crack to be thin (h<^L). Then the boundary conditions
on its surface can be applied to the corresponding segment of the #-axis. Thus
the crack is regarded as a line of discontinuity (in the xy-plane) on which the
normal component of the displacement u y = + \h is discontinuous.
Instead of h(x) we define a new unknown function p(x) by the formulae
L
h(x)= lp(x)dx, p(-x)=- P (x). (31.1)
x
The function p(x) may be conveniently, though purely formally, interpreted
as a density of straight dislocations lying in the s'-direction and continuously
distributed along the x-axis, with their Burgers vectors in the ^-direction.J
It has been shown in §27 that a dislocation line may be regarded as the edge of
a surface of discontinuity on which the displacement u has a discontinuity b.
In the form (31.1) the discontinuity h of the normal displacement at the point
f The quantitative theory of cracks discussed here is due to G. I. Barenblatt (1959).
% It is for this reason that the theory of cracks is described here in the chapter on dislocations,
although physically the phenomena are quite different.
§31 Equilibrium of a crack in an elastic medium 145
x is regarded as the sum of the Burgers vectors of all the dislocations lying to
the right of that point; the equation p{-x) = -p{x) signifies that the dis-
locations to the right and to the left of the point * = have opposite signs.
By means of this representation we can write down immediately an expression
for the normal stresses (a yy ) on the *-axis. These consist of the stresses
oyy {e) (x> 0) resulting from the external loads (which for brevity we denote by
p(x)) and the stresses o yy W{x) due to the deformation caused by the crack.
Regarding the latter stresses as being due to dislocations distributed over the
segment (-L, L), we obtain (similarly to (30.1))
ayrKx)= J[jm, (31.2)
j L i- x
for points in the segment ( - L, L) itself, the integral must be taken as a princi-
pal value. For an isotropic medium,
27r(l-cr) 4tt(1-<7 2 )
see §28, Problem 3. The stresses a xy due to such dislocations in an isotropic
medium are zero on the #-axis.
The boundary condition on the free surface of the crack, applied (as already
mentioned) to the corresponding segment of the *-axis, requires that the
normal stresses a yy = a 2/2/ < cr > +p(x) should be zero. This condition, however,
needs to be made more precise, for the following reason.
Let us make the assumption (which will be confirmed by the result) that
the edges of the crack join smoothly near its ends, so that the surfaces approach
very closely. Then it is necessary to take into account the forces of molecular
attraction between the surfaces ; the action of these forces extends to a distance
r large compared with interatomic distances. These forces will be of impor-
tance in a narrow region near the end of the crack where h<ro; the length of
this region will be denoted by d in order of magnitude, and will be estimated
later.
Let G be the force of molecular cohesion per unit area of the crack; it
depends on the distance h between the surfaces.! When these forces are taken
into account, the boundary condition becomes
a yy (rt+p(x)-G = 0. (31.4)
It is reasonable to suppose that the shape of the crack near its end is deter-
mined by the nature of the cohesion forces and does not depend on the external
loads applied to the body. Then, in finding the shape of the main part of the
crack from the external forces p(x), the quantity G becomes a given function
G(x) independent of p(x) (over the region d, outside which it is unimportant).
f In the macroscopic theory, the function G(x) is to be regarded as increasing smoothly, asL-x
decreases, up to a maximum value at the end of the crack.
6*
146 Dislocations §31
Substituting a yy ^) from (31.2) in (31.4), we thus obtain the following
integral equation for p(x) :
L Cp(i)d$ 1 1
P ^^ = -rftx)-^*) - «(*). (31-5)
J t — x D D
Since the ends of the crack are assumed not fixed, the stresses must remain
finite there. This means that, in solving the integral equation (31.5), we now
have the last of the cases discussed in §30, for which the solution is given by
(30.15). With the origin at the midpoint of the segment ( — L,L) this formula
becomes
pw= __ v(i2 _, 2) pj__y___. (31 . 6)
— Jj
The condition (30.14) must be satisfied, which in this case gives
l , , . l
C P( x ) dx _ C G(x)dx = q
(where the integrals from — L to L have been replaced by integrals from to
L, using the symmetry of the problem). Since G(x) is zero except in the range
L — x ~ d y in the second integral we can put L 2 — x 2, ^ 2L(L — x) ; the condition
(31.7) then becomes
p(x)dx M
FK } (31.8)
q V(L 2 -x*) V(2L)
where M denotes the constant
which depends on the medium concerned. This constant can be expressed in
terms of the ordinary macroscopic properties of the body, its elastic moduli
and surface tension a ; as will be shown later, the relation is
M=vT™£/(l-a 2 )]. (31.10)
The equation (31.8) determines the length 2L of the crack from the given
stress distribution p{x). For example, for a crack widened by concentrated
§31 Equilibrium of a crack in an elastic medium 147
forces /applied to the midpoints of the sides (p(x) =fB(x)) we find
2L=/ 2 /Af2
= /2(l-cT 2 )/7ra£'. (31.11)
It must be remembered, however, that stable equilibrium of a crack is not
possible for every distribution p(x). For instance, with uniform widening
stresses (p(x) = constant = po) (31.8) gives
1L = 4M 2 /t7 V
= 4a£/77(l-(72)/>o2. ( 31 - 12 )
This inverse relation (L decreasing when^o increases) shows that the state is
unstable. The value of L determined by (31.12) corresponds to unstable
equilibrium and gives the "critical" crack length: longer cracks grow spon-
taneously, but shorter ones close up, a result first derived by A. A. Griffith
(1920).
Let us now return to the consideration of the shape of the crack. WhenL -x<d,
the region L - 1 ~ d is the most important in the integral in (3 1 . 6) . The integral
can then be replaced by its limiting value as x^L ; the result is p = constant x
x \/(L — x), whencef
h(x) = constant x(L-xf 12 (L-x~d). (31.13)
We see that over the terminal region d the two sides of the crack in fact join
smoothly. The value of the coefficient in (31.13) depends on the properties of
the cohesion forces and can not be expressed in terms of the ordinary macro-
scopic parameters. J
For the part farther from the end, where d<L-x<^L, the region L-£~d
is again the most important in the integral in (31.6), and o>(£)^ -G(0I D - * n
addition to putting L 2 -x*z2L(L-x), L 2 -£ 2 z2L(L-£), we can here
replace g-x by L-x, obtaining /> = MJt^D^/{L-x), where M is the same
constant as in (31.9), (31.10). Hence
h(x) = 2MV(L-x)J7rW (d4L-x<L). (31.14)
Thus the end of the crack has a shape independent of the applied forces (and
therefore of the length of the crack) throughout the range L-x^L: when
L - x > d the shape is given by (3 1 . 14), and when L - x ~ d it has an infinitely
t In order to proceed to the limit we must first divide the integral in (31.6) into two integrals with
numerators a>(f) — m(L) and u(L); the second integral makes no contribution to the limiting value.
X An estimate of the coefficient in (3 1 . 1 3) gives a value of the order of Va/d, where a is the dimension
of an atom (using a ~ aE, M ~ EVa). An estimate of the length d is obtained from the condition
h{d) ~ ro, whence d ~ ro 2 /a > ro. It should be mentioned, however, that in practice the required
inequalities are satisfied only by a small margin, so that the resulting shape of the terminal projection
of the crack is not to be taken as exact.
148 Dislocations §31
sharp projection (31.13) (Fig. 28). The shape of the remainder of the crack
does depend on the applied forces.
Y*-*;.-
Fig. 28
If we ignore details, of the order of the radius of the action of the cohesion
forces, the crack therefore has a smooth outline with ends rounded according
to the parabolas (31.14), and this shape is entirely determined by the applied
forces and the ordinary macroscopic parameters. The small (~d) terminal
projections which actually occur are of fundamental significance, however,
since they ensure that the stresses remain finite at the ends of the crack.
The stresses caused by the crack on the continuation of the #-axis are given
by formula (31.2). At distances x—L such that d<^x—L<^L, we havej-
Oyy^(Tyy^)^MlTT^{x-L). (31.15)
The increase in the stresses as the edge of the crack is approached continues
according to this law up to distances x—L~d, and a vy then drops to zero at
the point x — L.
It remains to derive the formula (31.10) already given above, which relates
the constant M to the ordinary macroscopic quantities. To do this, we write
down the condition for the total free energy to be a minimum by equating to
zero its variation under a change in the length L.
Firstly, when the length of the crack increases by SL the surface energy at
its two free surfaces increases by &F S urf = 2aSL. Secondly, the "opening" of
the crack end reduces the elastic energy F e \ by %$oyy(x)r)(x)dx, where rj(x) is
the difference in width between the displaced and undisplaced crack shapes.
Since the shape of the crack end is independent of its length, r)(x) = h(x — 8L) —
— h{x). The stress a yy = for x<L, and h(x) = for x> L. Hence
L+SL
8F e i= — § \o yy (x)h(x — 8L)dx.
f The integral is easily calculated directly, but it is not necessary to do this if we use the relation
between the functions p(x) for x < L and a ay (cr) for x > L, which is evident from the results of §30.
§31 Equilibrium of a crack in an elastic medium 149
Substituting (31.14) and (31.15), we find
L+dL
M 2 r iL+SL-x,
L
8L
M 2 r Vydy
M *SL.
Finally, the condition SF^t+BFei = gives the relation M 2 = 47r 2 oD, and
hence we have (31.10).|
t It may be noted that the theory described above, including the relation (31.10), is in fact applicable
as it stands only to ideally brittle bodies, i.e. those which remain linearly elastic up to fracture, such as
glass and fused quartz. In bodies which exhibit plasticity the formation of the crack may be accompanied
by plastic deformation at its ends.
CHAPTER V
THERMAL CONDUCTION AND VISCOSITY IN SOLIDS
§32. The equation of thermal conduction in solids
Non-uniform heating of a solid does not cause convection as it generally
does in fluids. Hence the transfer of heat is effected in solids by thermal
conduction alone. The processes of thermal conduction in solids are there-
fore described by somewhat simpler equations than those for fluids, where
they are complicated by convection.
The equation of thermal conduction in a solid can be derived immediately
from the law of conservation of energy in the form of an "equation of con-
tinuity for heat". The amount of heat absorbed per unit time in unit volume
of the body is TdS/dt, where S is the entropy per unit volume. This must
be put equal to — div q, where q is the heat flux density. This flux can
almost always be written as — k grad T, i.e. it is proportional to the tempera-
ture gradient (k being the thermal conductivity). Thus
TdSJdt = div(/c grad T). (32.1)
According to formula (6.4), the entropy can be written as
S = S (T) + K<x.u iiy
where a is the thermal expansion coefficient and So the entropy in the
undeformed state. We shall suppose that, as usually happens, the tempera-
ture differences in the body are so small that quantities such as k, a, etc.
may be regarded as constants. Then equation (32.1), after substitution of
the above expression for S, becomes
dSn dtiu
T— + olKT— = *AT.
dt dt
According to a well-known formula of thermodynamics, we have
C p —C v = KofiT,
whence
olKT = (C p —C v )l<x.
The time derivative of So can be written as (dSo/dT) • (dT/dt), where the
derivative dSojdT is taken for uu = div u = 0, i.e. at constant volume, and
therefore is equal to CJT.
150
§32 The equation of thermal conduction in solids 151
The resulting equation of thermal conduction is
C ?— + Cv - Cv - divu = kAT. (32.2)
dt a dt
In order to obtain a complete system of equations, it is necessary to add an
equation describing the deformation of a non-uniformly heated body. This is
the equilibrium equation (7.8) :
2(1 _ a ) grad div u- (1 - 2a) curl curl u = |a(l + a) grad T. (32.3)
From equation (32.3) we can in principle determine the deformation of the
body for any given temperature distribution. Substituting the expression for
divu thus obtained in equation (32.2), we derive an equation giving the
temperature distribution, in which the only unknown function is T (x, y, z, t).
For example, let us consider thermal conduction in an infinite solid in
which the temperature distribution satisfies only one condition: at infinity,
the temperature tends to a constant value T , and there is no deformation.
In such a case equation (32.3) leads to the following relation between div u
and T (see §7, Problem 8) :
1 + cr
divu = — -a(T-To).
3(1 - a)
Substituting this expression in (32.2), we obtain
(l + a)C p + 2(l-2a)C v W_ =KATy (32 . 4)
3(1 -a) dt
which is the ordinary equation of thermal conduction.
An equation of this type also describes the temperature distribution along
a thin straight rod, if one (or both) of its ends is free. The temperature may
be assumed constant over any transverse cross-section, so that T is a function
only of the co-ordinate x along the rod and of the time. The thermal expan-
sion of such a rod causes a change in its length, but no departure from straight-
ness and no internal stresses. Hence it is clear that the derivative dSjdt in
the general equation (32.1) must be taken at constant pressure and, since
(dSldt) p = Cp/T, the temperature distribution will satisfy the one-dimen-
sional thermal conduction equation C v dTjdt = Kd 2 T\dx 2 .
It should be mentioned, however, that the temperature distribution m a
solid can in practice always be determined, with sufficient accuracy, by a
simple thermal conduction equation. The reason is that the second term on
the left-hand side of equation (32.2) is a correction of order {C P ~C V )IC V
relative to the first term. In solids, however, the difference between the two
specific heats is usually very small, and if it is neglected the equation of
thermal conduction in solids can always be written
BTIft = x* T > ( 32 ' 5)
152 Thermal Conduction and Viscosity in Solids §33
where x is the thermometric conductivity, defined as the ratio of the ther-
mal conductivity k to some mean specific heat per unit volume C.
§33. Thermal conduction in crystals
In an anisotropic body, the direction of the heat flux q is not in general
that of the temperature gradient. Hence, instead of the formula
q = _ K grad T
relating q to the temperature gradient, we have in a crystal the more general
relation
qt = -K tk dT/dx k . (33.1)
The tensor K ik) of rank two, is called the thermal conductivity tensor of the
crystal. In accordance with (33.1), the equation of thermal conduction (32.5)
has also a more general form,
dT d*T
c ir - * w < 33 - 2 >
A general theorem can be stated: the thermal conductivity tensor is
symmetrical, i.e.
Kik = K ki - (33.3)
This relation, which we shall now prove, is a consequence of the symmetry
of the kinetic coefficients.f
The rate of increase of the total entropy of the body by irreversible pro-
cesses of thermal conduction is
C div q C Q C 1
St0t = " J T dV = " J div Y dF+ J q-grad-dF.
The first integral, on being transformed into a surface integral, is seen to be
zero. Thus
= Jq.grad— dV = - j
Stot = | q-grad^dF = - | q '**£ T dV,
or
1 dT
ito,= -Jr^ dF - < 33 - + >
In accordance with the general definition of the kinetic coefficients,:!: we
t See Statistical Physics, §122.
J We here use the definition given in Fluid Mechanics, §58.
34 Viscosity of solids 153
can deduce from (33.4) that in the case considered the coefficients T 2 K tk in
1 dT
/l dT\
are kinetic coefficients. Hence the result (33.3) follows immediately from the
symmetry of the kinetic coefficients.
The quadratic form
dT dT dT
dxi dxt dxjc
must be positive, since the time derivative (33.4) of the entropy must be
positive. The condition for a quadratic form to be positive is that the eigen-
values of the matrix of its coefficients are positive. Hence all the principal
values of the thermal conductivity tensor K ilc are always positive; this is
evident also from simple considerations regarding the direction of the heat
flux.
The number of independent components of the tensor km depends on the
symmetry of the crystal. Since the tensor K ik is symmetrical, this number is
evidently the same as the number for the thermal expansion tensor (§10),
which is also a symmetrical tensor of rank two.
§34. Viscosity of solids
In discussing motion in elastic bodies, we have so far assumed that the
deformation is reversible. In reality, this process is thermodynamically
reversible only if it occurs with infinitesimal speed, so that thermodynamic
equilibrium is established in the body at every instant. An actual motion,
however, has finite velocities ; the body is not in equilibrium at every instant,
and therefore processes will take place in it which tend to return it to equili-
brium. The existence of these processes has the result that the motion is
irreversible, and, in particular, mechanical energyf is dissipated, ultimately
into heat.
The dissipation of energy occurs by two means. Firstly, when the tempera-
ture at different points in the body is different, irreversible processes of thermal
conduction take place in it. Secondly, if any internal motion occurs in the
body, there are irreversible processes arising from the finite velocity of
that motion. This means of energy dissipation may be referred to, as in
fluids, as internal friction or viscosity.
In most cases the velocity of macroscopic motions in the body is so small
that the energy dissipation is not considerable. Such "almost irreversible"
processes can be described by means of what is called the dissipative function.%
f By mechanical energy we here mean the sum of the kinetic energy of the macroscopic motion in
the elastic body and its (elastic) potential energy arising from the deformation.
% See Statistical Physics, §123.
154 Thermal Conduction and Viscosity in Solids §34
If we have a mechanical system whose motion involves the dissipation of
energy, this motion can be described by the ordinary equations of motion,
with the forces acting on the system augmented by the dissipative forces or
frictional forces, which are linear functions of the velocities. These forces
can be written as the velocity derivatives of a certain quadratic function T
of the velocities, called the dissipative function. The frictional force f a
corresponding to a generalised co-ordinate q a of the system is then given by
fa — - WlHa' The dissipative function T is a positive quadratic form in
the velocities q a . The above relation is equivalent to
SY = ~2/«%> (34.1)
a
where ST is the change in the dissipative function caused by an infinitesimal
change in the velocities. It can also be shown that the dissipative function is
half the decrease in the mechanical energy of the system per unit time.
It is easy to generalise equation (34.1) to the case of motion with friction
in a continuous medium. The state of the system is then defined by a con-
tinuum of generalised co-ordinates. These are the displacement vector u at
each point in the body. Accordingly, the relation (34.1) can be written in
the integral form
SJTdF= - jfiSutdV, (34.2)
where ft are the components of the dissipative force vector f per unit volume
of the body; we write the total dissipative function for the body as JT dV,
where T is the dissipative function per unit volume.
Let us now determine the general form of the dissipative function T for
deformed bodies. The function T, which describes the internal friction,
must be zero if there is no internal friction, and in particular if the body
executes only a general translatory or rotary motion. In other words, the dis-
sipative function must be zero if u = constant or u = SlXr. This means that it
must depend not on the velocity itself but on its gradient, and can contain only
such combinations of the derivatives as vanish when u = SI X r. These are
the sums
dxjc dxi
i.e. the time derivatives u\u of the components of the strain tensor, f Thus
the dissipative function must be a quadratic function of w^. The most
general form of such a function is
^ = IViklmUikUim- (34.3)
t Cf. the entirely analogous arguments on the viscosity of fluids in Fluid Mechanics, §15.
§35 The absorption of sound in solids 155
The tensor rj ik i mt of rank four, may be called the viscosity tensor. It has the
following evident symmetry properties:
T)iklm = Vlmik = f]kilm = 7 ]ikmh ("•*)
The expression (34.3) is exactly analogous to the expression (10.1) for the
free energy of a crystal: the elastic modulus tensor is replaced by the tensor
rjikim, and u ik by u ik . Hence the results obtained in §10 for the tensor X ik i m
in crystals of various symmetries are wholly valid for the tensor r]m m also.
In particular, the tensor rjmm in an isotropic body has only two independent
components, and Y can be written in a form analogous to the expression
(4.3) for the elastic energy of an isotropic body:
Y = ^ite- ^imif+mi 2 , (34.5)
where rj and £ are the two coefficients of viscosity. Since T is a positive
function, the coefficients r\ and £ must be positive.
The relation (34.2) is entirely analogous to that for the elastic free energy,
8 $FdV = -IFihm dV, where F t = da ik \dx k is the force per unit volume.
Hence the expression for the dissipative force ft in terms of the tensor u ik
can be written down at once by analogy with the expression for F t in terms
of ui k . We have
fi = da'ikldxic, (34.6)
where the dissipative stress tensor a'a is defined by
a'ijc = BYjdUiJc = rjmmUim. (34.7)
The viscosity can therefore be taken into account in the equations of motion
by simply replacing the stress tensor <j ik in those equations by the sum
oi ic + cr'i k .
In an isotropic body,
a'i k = 27](ui k -l8i k uu) + ^uiiB ik . (34.8)
This expression is, as we should expect, formally identical with that for the
viscosity stress tensor in a fluid.
§35. The absorption of sound in solids
The absorption of sound in solids can be calculated in a manner entirely
analogous to that used for fluids.f Here we shall give the calculations for an
isotropic body. The thermal-conduction part of the energy dissipation
£nech is given by the integral -(k/T)/ (grad Tf dV. On account of viscosity,
an amount of energy 2Y is dissipated per unit time and volume, so that the
total viscosity part of E m ech is -2 J Y dV. Using the expression (34.5), we
t See Fluid Mechanics, §77.
156 Thermal Conduction and Viscosity in Solids §35
therefore have
imecn = - j /(grad Tf dV-2r)j(ui k -ls8 ik my dV-^ju^ dV. (35.1)
To calculate the temperature gradient, we use the fact that sound oscilla-
tions are adiabatic in the first approximation. Using the expression (6.4) for
the entropy, we can write the adiabatic condition as So(T) + Koaiu = So(Tq),
where To is the temperature in the undeformed state. Expanding the differ-
ence So(T) — So(To) in powers of T—To, we have as far as the first-order
terms S (T)-So(T ) = (T-T ) dSo/dT = C v {T-T )jT . The derivative
of the entropy is taken for uu = 0, i.e. at constant volume. Thus
T— T = — T(x.KuujC v .
Using also the relations K = Kiso = C V K^\C V and K a a/p = cp-\c?\?>,
we can rewrite this result as
T*p(c?-4cflZ)
T-T = — uu. (35.2)
Let us first consider the absorption of transverse sound waves. The
thermal conduction cannot result in the absorption of these waves (in the
approximation considered). For, in a transverse wave, we have uu — 0, and
therefore the temperature in it is constant, by (35.2). Let the wave be propa-
gated along the #-axis; then
u x = 0, u y = uoycos(kx—cot), u z = uo z cos(kx—cot),
and the only non-zero components of the deformation tensor are
u X y = — \kuoy sin(&K— exit), u X z = — \hioz sin(A#— cot).
We shall consider the energy dissipation per unit volume of the body ; the
(time) average value of this quantity is, from (35.1),
^mech = -ii7 ft)4 ( M Oy 2 + Moz 2 )M 2 ,
where we have put k = co/ct. The total mean energy of the wave is twice
the mean kinetic energy, i.e.
E = pjtfdV;
for unit volume we have
£ = |/oco 2 (woy 2 + «oz 2 )-
The sound absorption coefficient is defined as the ratio of the mean energy
dissipation to twice the mean energy flux in the wave; this quantity gives
the manner of variation of the wave amplitude with distance. The amplitude
§35 The absorption of sound in solids 157
decreases as e~ y *. Thus we find the following expression for the absorption
coefficient for transverse waves:
yt = illmecul/^fi = -n^Ppcf. (35.3)
In a longitudinal sound wave u z - uocos(kx-cot), u y = u z = 0. A
similar calculation, using formulae (35.1) and (35.2), gives
CO
[&♦«)« -sn- <■">
71 ~2 P c?
These formulae relate, strictly speaking, only to a completely isotropic and
amorphous body. They give, however, the correct order of magnitude for
the absorption of sound in anisotropic single crystals also.
The absorption of sound in polycrystalline bodies exhibits peculiar proper-
ties. If the wavelength A of the sound is small in comparison with the
dimensions a of the individual crystallites, then the sound is absorbed in
each crystallite in the same way as in a large crystal, and the absorption
coefficient is proportional to a> 2 .
If A > «, however, the nature of the absorption is different. In such a
wave we can assume that each crystallite is subject to a uniformly distributed
pressure. However, since the crystallites are anisotropic, and so are the
boundary conditions at their surfaces of contact, the resulting deformation is
not uniform. It varies considerably (by an amount of the same order as
itself) over the dimension of a crystallite, and not over one wavelength as in a
homogeneous body. When sound is absorbed, the rates of change of the
deformation (u ik ) and the temperature gradients are of importance. Of
these, the former are still of the usual order of magnitude. The temperature
gradients within each crystallite are anomalously large, however. Hence the
absorption due to thermal conduction will be large compared with that due to
viscosity, and only the former need be calculated.
Let us consider two limiting cases. The time during which the temperature
is equalised by thermal conduction over distances ~ a (the relaxation time
for thermal conduction) is of the order of a 2 )x- Let us first assume that
<o ^ xl<* 2 - Tnis means that the relaxation time is small compared with the
period of the oscillations in the wave, and so thermal equilibrium is nearly
established in each crystallite; in this case we have almost isothermal oscilla-
tions.
Let T be the temperature difference in a crystallite, and T ' the corres-
ponding difference in an adiabatic process. The heat transferred by thermal
conduction per unit volume is -divq = k[\T ~ kT/o 2 . The amount of
heat evolved in the deformation is of the order of Tq'C ~ ojTo'C, where
C is the specific heat. Equating the two, we obtain T ~ To'oja 2 jx- The
temperature varies by an amount of the order of T over the dimension of
the crystallite, and so its gradient is of magnitude ~ T'\a. Finally, TV is
158 Thermal Conduction and Viscosity in Solids §35
found from (35.2), with uu ~ ku ~ coufc (u being the amplitude of the
displacement vector):
T ' ~ Txpccou/C] (35.5)
in obtaining orders of magnitude, we naturally neglect the difference between
the various velocities of sound. Using these results, we can calculate the
energy dissipated per unit volume :
- K K (T'\ 2
£ m ax~y(gradT)2~-^-j.
Dividing this by the energy flux cE ~ cpa> 2 u 2 , we find the damping coefficient
to be
y ~ TaL 2 pca 2 a) 2 lxC for co < x/a 2 (35.6)
(C. Zener 1938). Comparing this expression with the general expressions
(35.3) and (35.4), we can say that, in the case considered, the absorption of
sound by a polycrystalline body is the same as if it had a viscosity
which is much larger than the actual viscosity of the component crystallites.
Next, let us consider the opposite limiting case, where o> > xl a2 - I* 1 other
words, the relaxation time is large compared with the period of oscillations
in the wave, and no noticeable equalisation of the temperature differences
due to the deformation can occur in one period. It would be incorrect,
however, to suppose that the temperature gradients which determine the
absorption of sound are of the order of To' /a. This assumption would take
into account only thermal conduction in each crystallite, whereas heat ex-
change between neighbouring crystallites must be of importance in the case in
question (M. A. Isakovich 1948). If the crystallites were thermally insulated
the temperature differences occurring at their boundaries would be of the
same order To' as those within each individual crystallite. In reality, however,
the boundary conditions require the continuity of the temperature across
the surface separating two crystallites. We therefore have "temperature
waves" propagated away from the boundary into the crystallite; these are
damped at a distance^ 8 ~ vXx/ 60 )- I* 1 tne case under consideration 8 <^ a,
i.e. the main temperature gradient is of the order of To'/8 and occurs over
distances small compared with the total dimension of a crystallite. The cor-
responding fraction of the volume of the crystallite is ~ a 2 8 ; taking the ratio
t It may be recalled that, if a thermally conducting medium is bounded by the plane x — 0, at
which the excess temperature varies periodically according to T" = T 'e r ~* tt) ', then the temperature
distribution in the medium is given by the "temperature wave" T" = T^e-i^t e-(l+t)xv / (««'/2x); see
Fluid Mechanics, §52.
§35 The absorption of sound in solids 159
of this to the total volume - a 3 , we find the mean energy dissipation
Emech ~ ~f\T) a* ~ Ta8 '
Substituting for T ' the expression (35.5) and dividing by cE ~ cpuPu\ we
obtain the required absorption coefficient:
y ~ T<x. 2 pcV( X co)laC for co > x/a 2 . (35.7)
It is proportional to the square root of the frequency-!
Thus the sound absorption coefficient in a polycrystalline body varies as
a? at very low frequencies (co <^ x /« 2 )*> for xl<* 2 < °> < c l a [t varies as V w »
and for a> > c/a it again varies as a> 2 .
Similar considerations hold for the damping of transverse waves in thin
rods and plates (C. Zener 1938). If h is the thickness of the rod or plate,
then for A > h the transverse temperature gradient is important, and the
damping is mainly due to thermal conduction (see the Problems). If also
o> ^ x jh 2 , the oscillations may be regarded as isothermal, and therefore, in
determining (for example) the characteristic frequencies of vibrations of the
rod or plate, the isothermal values of the moduli of elasticity must be used.
PROBLEMS
Problem 1. Determine the damping coefficient for longitudinal vibrations ofa rod. _
Solution. The damping coefficient for the vibrations is denned as £ = |£mech|/2-E;
the amplitude of the vibrations diminishes with time as e'fiK
In a longitudinal wave, any short section of the rod is subject to simple extension or com-
pression; the components of the strain tensor are u zz = du z Idz, u xx = u yy = —o^duzjdz.
We put u z = Mo cos kz cos wt, where k = «>/ V(#ad/p)- Calculations similar to those given
above lead to the following expression for the damping coefficient:
Here we have written £ a d and <* a d in terms of the velocities c u c t by means of formulae (22.4).
Problem 2. The same as Problem 1, but for longitudinal oscillations of a plate.
Solution. For waves whose direction of oscillation is parallel to that of their propagation
(the w-axis, say) the non-zero components of the strain tensor are
U X x = dUxfix, U ZZ = -[o-ad/tl-o-acOFM*/ 9 *;
see (13.1). The velocity of propagation of these waves is \/[£'ad/p(l - ff ad 2 )]- A calculation
gives
_ afito 3cfi + 4c£-6c?c? W KT^p z (l + a & a) 2 \
P = Tp[Z cM{&-<*) + <#(<#- <$) + 9C v 2 ''
t The same frequency dependence is found for the absorption of sound propagated in a fluid near
a solid wall (in a pipe, for instance); see Fluid Mechanics, §77, Problems.
160 Thermal Conduction and Viscosity in Solids §35
For waves whose direction of oscillation is perpendicular to the direction of propagation,
«n = 0, and the damping is caused only by the viscosity 17. In this case the damping coeffi-
cient is fi = r)w*l2pct*. This applies also to the damping of torsional vibrations of rods.
Problem 3. Determine the damping coefficient for transverse vibrations of a rod (with
frequencies such that a> > x/h', where h is the thickness of the rod).
Solution. The damping is due mainly to thermal conduction. According to §17, we have
for each volume element in the rod Uzz = x/R, u xx = u yy = —a^xIR (for bending in the
*ar-plane) ; for at > x/h', the vibrations are adiabatic. For small deflections the radius of
curvature R = \\X", so that u ti = (1 — 2o & a)xX", the prime denoting differentiation with
respect to z. The temperature varies most rapidly across the rod, and so (grad T) J
«(3773*)*. Using (35.1) and (35.2), we obtain for the total mean energy dissipation in the
rod — (KTa*E & &*SI9C v *) / X"* dz, where S is th e cros s-sectional area of the rod. The mean
total energy is twice the potential energy E^l y \X" % dz. The damping coefficient is
j8 = KT<x?SE aa /l8IyC p Z.
Problem 4. The same as Problem 3, but for transverse vibrations of a plate.
Solution. According to (11.4), we have for any volume element in the plate
l-2a ad BH
m = _ z
1 — o"ad OX*
for bending in the roar-plane. The energy dissipation is found from formulae ( 35 .1) and (5.2)
and the mean total energy is twice the expression (11.6). The damping coefficient is
2 K T^E &a 1 + gad _ 2kTv* p (3^-4^)2^2
P ~~ ZC P W 1-crad " 3C,«A« ' W-cflcfi '
Problem 5. Determine the change in the characteristic frequencies of transverse vibrations
of a rod due to the fact that the vibrations are not adiabatic. The rod is in the form of a
long plate of thickness h. The surface of the rod is supposed thermally insulated.
Solution. Let T & &(x, t) be the temperature distribution in the rod for adiabatic vibra-
tions, and T(x, t) the actual temperature distribution; x is a co-ordinate across the thickness
of the rod, and the temperature variation in the yz-ptene is neglected. Since, for T = T a d,
there is no heat exchange between various parts of the body, it is clear that the thermal con-
duction equation must be
d 3 2 r
u {T - T ** ) = x w-
For periodic vibrations of frequency a>, the differences r a( j = T a d~" "^o» f — T—T from
the equilibrium temperature T are proportional to e~ iM , and we have r" -\-imr\x = i<^T&Cilx>
the prime denoting differentiation with respect to x. Since, by (35.2), r a( j is proportional
to un, and the components «,# are proportional to x (see §17), it follows that T a d = Ax, where
A is a constant which need not be calculated, since it does not appear in the final result. The
solution of the equation T"+iair/x = iwAx/x, with the boundary condition t' = for
x = ±ih( the surface of the rod being insulated), is
(Sill nX \
*~i — in- ' k = (i+iWH2 x ).
K COS ~hnfl /
The moment M y of the internal stress forces in a rod bent in the xsr-plane is composed of
the isothermal part M y< i s0 (i.e. the value for isothermal bending) and the part due to the
§36 Highly viscous fluids 161
non-uniform heating of the rod. If M„, a d is the moment in adiabatic bending, the second
part of the moment is reduced from My.^a— M v ,iao in the ratio
hh ift
1 +f(o>) = j zr dz/j zr & a dz.
-ift -ih
Defining the Young's modulus # w for any frequency a» as the coefficient of proportionality
between M„ and IJR (see (17.8)), and noticing that E^-E = E l Ta*l9C 9 (see (6.8); E is
the isothermal Young's modulus), we can put
E^E+ll+fMWTotpCj,.
A calculation shows that /(o>) = (24/#V» 8 )(ita -tan \kh). For a. -> oo we obtain / = 1,
which is correct, since Eoo = £ad» and for <a -*■ 0, / = and E = E.
The frequencies of the characteristic vibrations are proportional to the square root of the
Young's modulus (see §25, Problems 4-6). Hence
"="o[l+/(*>o)— ],
where a» are the characteristic frequencies for adiabatic vibrations. This value of to is
complex. Separating the real and imaginary parts (<o = o'+tf), we find the characteristic
frequencies
r ETa? 1 sinh£-sin£l
"' = °T " ~3C^'T*' cosh £+ cos d
and the damping coefficient
2ETa?x\ 1 sinh^+sinn
^ = 3CpA2 [ ~ i'coshl+cosd'
where £ = A\/( a> o/2x)«
For large £ the frequency a> tends to w , as it should, and the damping coefficient to
2£roc 2 x/3C,A 2 , in accordance with the result of Problem 3.
Small values of £ correspond to almost isothermal conditions ; in this case
co ^ coo\ 1 - 1S „ ) « cooV(EIE a d),
and the damping coefficient /3 = i?T , a2A 2 w 2 /180C,X'
§36. Highly viscous fluids
For typical fluids, the Navier-Stokes equations are valid if the periods of
the motion are large compared with times characterising the molecules. This,
however, is not true for very viscous fluids. In such fluids, the usual equations
of fluid mechanics become invalid for much larger periods of the motion.
There are viscous fluids which, during short intervals of time (though these
are long compared with molecular times), behave as solids (for instance,
glycerine and resin). Amorphous solids (for instance, glass) may be regarded
as a limiting case of such fluids having a very large viscosity.
The properties of these fluids can be described by the following method,
due to Maxwell. They are elastically deformed during short intervals of time.
162 Thermal Conduction and Viscosity in Solids §36
When the deformation ceases, shear stresses remain in them, although these
are damped in the course of time, so that after a sufficiently long time almost
no internal stress remains in the fluid. Let r be of the order of the time during
which the stresses are damped (sometimes called the Maxwellian relaxation
time). Let us suppose that the fluid is subjected to some variable external forces,
which vary periodically in time with frequency w. If the period l/a» is large
compared with the relaxation time t, i.e. wr <^ 1, the fluid under consideration
will behave as an ordinary viscous fluid. If, however, the frequency co is suffi-
ciently large (so that cot > 1), the fluid will behave as an amorphous solid.
In accordance with these "intermediate" properties, the fluids in question
can be characterised by both a viscosity coefficient r\ and a modulus of
rigidity \l. It is easy to obtain a relation between the orders of magnitude of
rj, fi and the relaxation time r. When periodic forces of sufficiently small
frequency act, and so the fluid behaves like an ordinary fluid, the stress tensor
is given by the usual expression for viscosity stresses in a fluid, i.e.
o'ik = ^]Uik = —licorjuijc.
In the opposite limit of large frequencies, the fluid behaves like a solid, and
the internal stresses must be given by the formulae of the theory of elasticity,
i.e. one = 2[m.ik\ we are speaking of pure shear deformations, i.e. we assume
that ua — an = 0. For frequencies co ~ 1/r, the stresses given by these
two expressions must be of the same order of magnitude. T1ius?7m/At ~ /x-m/A,
whence
7] ~ T[Jb. (36.1)
This is the required relation.
Finally, let us derive the equation of motion which qualitatively describes
the behaviour of these fluids. To do so, we make a very simple assumption
concerning the damping of the internal stresses (when motion ceases):
namely, that they are damped exponentially, i.e. dcr^/d* = — ct^/t. In a
solid, however, we have o% = 2/xm^, and so daacjdt = Ifidutjc/dt. It is easy
to see that the equation
da ilc a i1c dune
= Zu. (3o.Z)
dt T ^ dt ^ '
gives the correct result in both limiting cases of slow and rapid motions, and
may therefore serve as an interpolatory equation for intermediate cases.
For example, in periodic motion, where uiu and cruc depend on the time
through a factor e~ iwt , we have from (36.2) —ioianc + oadr = —2iu)fj,Uijc,
whence
0ik = - — - — • (36.3)
1 + l/COT
For cot > 1, this formula gives aa = I^um, i.e. the usual expression for
solid bodies, while for cut <^ 1 we have ow = —2ioi[irUik = 2/x,t«m, the
usual expression for a fluid of viscosity /jlt.
INDEX
Absorption of sound
in polycrystalline bodies 157-159
in solids 155-161
Adiabatic moduli 16-17
Anharmonic effects 119-122
Bending
energy 58, 62—65
moment 78
of plates 44-53, 58-62
principal planes of 78
of rods 75-97
of shells 62-68
strip 64
waves 114-115
Biharmonic equation 18, 24
Bulk modulus 1 1
Burgers vector 124
Coefficient
of compression 12
of extension 1 3
of hydrostatic compression 12
of thermal expansion 16
in crystals 42
of unilateral compression 1 5
Combination frequencies 120
Compression
modulus of 11
of rods 13-15
uniform (hydrostatic) 6, 10
coefficient of 12
modulus of 11
unilateral 1 5
coefficient of 15
Contact problems 30-37
Crack in elastic medium 144-149
Crystals
biaxial 42
elastic moduli of 37—43
elastic properties of 37-43
elastic waves in 106-108
free energy of 37-41
thermal conduction in 152-153
thermal expansion of 42
uniaxial 42
Damping, see Absorption
Deformation 1
Deformation (continued)
adiabatic 16-17, 101
on contact 30-37
due to dislocation 123-131
elastic 8
homogeneous 1 3-1 5
of infinite medium 29
isothermal 16-17
plane 19-20, 24
plastic 8, 135
residual 8
of shells 62-68
with change of temperature 15-17
thermodynamic relation for 9
thermodynamics of 8-9
torsional 69
Dislocations (IV) 123-149
continuous distribution of 134-139
density tensor 134
edge 123
flux density tensor 137
interacting 139-143
moment tensor 127
polarisation 138
screw 124
sign of 124n.
in stress field 131-134
wall 131
Displacement vector 1
Dissipative
forces 1 54
function 153
Distortion tensor 125
Elastic
instability 97
line 77
modulus tensor 37 ; see also Modulus
oscillations, see Elastic waves
plane 53
waves (III) 101-122
absorption of 155-161
anharmonic 1 1 9-1 22
bending 114-115
in crystals 106-108
damping of 155-161
in isotropic media 101-106
longitudinal 102, 113-114
Rayleigh 109-113
reflection and refraction of 103-105
in rods and plates 113-118
163
164
Index
Elastic waves (continued)
surface 109-113
torsional 115-116
transverse 102, 114-115
Energy
bending 58
for shell 62-65
stretching 58
for shell 62-65
see also Free energy-
Equilibrium
equations of 7
for infinite medium 29
for isotropic bodies 18
for membranes 61
for plates 49, 54, 60, 61
for rods 82-83, 85, 89
of elastic half-space 25-29
of isotropic bodies 17-25
of rods and plates (II) 44-100
Extension
coefficient of 13
modulus of 13
of rods 13-15
Free energy
of deformed body 1 0-1 1 , 1 6 , 1 22
of deformed crystal 37-41
elastic lln.
of bent plate 46, 59
of bent rod 77, 80
of unilaterally compressed rod 1 5
of stretched rod 14
of twisted rod 72, 80
Frictional forces 154
Group velocity of waves 107
Hooke's law 12
Hydrostatic compression 6, 10
coefficient of 12
modulus of 11
Isothermal moduli 1 6-1 7
Modulus (continued)
of compression 1 1
elastic
for crystals 37-43
for polycrystalline bodies 41-42
of extension 1 3
of hydrostatic compression 1 1
isothermal 16-17
of rigidity 1 1
shear 1 1
Young's 13
for cubic crystals 43
Neutral surface
Notation viii
44,75
49
Plates
bending of 44-53, 58-62
equation of equilibrium for
clamped 49-50
large deflections of 58-62
longitudinal deformations of 53-57
supported 51
thin 44
vibration of 113-118,159
Poisson's ratio 13
Rayleigh waves 109-113
Reflection and refraction of elastic waves
103-105
Rigidity
cylindrical 49n.
flexural 49n., 90
modulus of 11
torsional 72
Rods
bending of 75-97
clamped 83
equations of equilibrium for 82-83, 85,
89
extension and compression of 1 3-1 5
hinged 84
small deflections of 89-97
supported 84
torsion of 68-75, 78-97
vertical, deformation of 20
vibration of 113-118, 159-161
Lam6 coefficients 10
Maxwellian relaxation time 162
Membrane 61, 63
Modulus
adiabatic 16-17
bulk 11
Shear
modulus 1 1
pure 10
Shearing force 89
Shells 62
deformation of 62-68
Slip plane 133
Stability of elastic systems
65, 97-100
Index
165
Strain tensor 2-4
in cylindrical co-ordinates 4
diagonalisation of 2
principal axes of 2
principal values of 2
in spherical co-ordinates 3
in terms of stress tensor 12, 14, 15
Stress function 20, 54
Stress, plane 54n.
Stress tensor 5-7, 11
mean value of 7
in terms of strain tensor 11 , 14, 37
Stresses
concentration of 25, 57
internal 4
moment of 5-6
Strings 92
Summation rule In.
Surface waves 109-113
Tensor ellipsoid 42
Theory of elasticity 1
fundamental equations of (I) 1-43
Thermal conduction
in crystals 152-153
in solids 150-153
Thermal conductivity tensor 152
Thermal expansion 15-17
in crystals 42
Thermal expansion coefficient 16
in crystals 42
Torsion
angle 69, 79
function 69
of rods 68-75,78-97
Torsional
rigidity 72
vibration 115-116
Velocity of sound
longitudinal 102
transverse 102
Vibration
anharmonic 119-122
of rods and plates 113-118
torsional 115-116
Viscosity
high, in fluids 161-162
of solids 153-155
tensor 155
Waves, see Elastic waves
Young's modulus 13
for cubic crystals 43
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